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Ch. 1 Quiz: Stoichiometry.
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Question 1 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
1. Question
1. How many oxygen atoms are present in 88 grams of CO_{2}?Correct / You marked this questionIt is often helpful to work backwards when solving stoichiometric problems. In order to determine how many oxygen atoms are present in this sample, we must know how many moles of oxygen there are. We can’t determine how many moles of oxygen there are until we know how many moles of CO2 we have. We can calculate how many moles of CO_{2} we have by using its molecular weight: each mole of CO_{2} weighs 12g + 16g + 16g = 44g (the 12g comes from C having a mass of 12 g/mol, and the 16g comes from O having a mass of 16g/mol). Thus, we can write:
88g of CO_{2} x (^{1 mol CO2}/_{44g of CO}_{2}) = 2 moles of CO_{2}
Each mole of CO_{2} contains 2 moles of oxygen. That means we must have:
2 moles of CO_{2} x (^{2 moles of oxygen}/_{1 mole of CO}_{2}) = 4 moles of oxygen
Each mole of oxygen contains 6.022 x 10^{23} oxygen atoms:
4 moles of oxygen x (^{6.022 x 10}^{23} oxygen atoms/_{1 mole of oxygen}) = 24 x 10^{23} oxygen atoms
We can do the above math in our head by simply rounding 6.022 down to 6 and multiplying 6 by 4, then tacking on x 10^{23}. Rewriting our answer in proper scientific notation, we get 2.4 x 10^{24} oxygen atoms.
Incorrect / You marked this questionIt is often helpful to work backwards when solving stoichiometric problems. In order to determine how many oxygen atoms are present in this sample, we must know how many moles of oxygen there are. We can’t determine how many moles of oxygen there are until we know how many moles of CO2 we have. We can calculate how many moles of CO_{2} we have by using its molecular weight: each mole of CO_{2} weighs 12g + 16g + 16g = 44g (the 12g comes from C having a mass of 12 g/mol, and the 16g comes from O having a mass of 16g/mol). Thus, we can write:
88g of CO_{2} x (^{1 mol CO2}/_{44g of CO}_{2}) = 2 moles of CO_{2}
Each mole of CO_{2} contains 2 moles of oxygen. That means we must have:
2 moles of CO_{2} x (^{2 moles of oxygen}/_{1 mole of CO}_{2}) = 4 moles of oxygen
Each mole of oxygen contains 6.022 x 10^{23} oxygen atoms:
4 moles of oxygen x (^{6.022 x 10}^{23} oxygen atoms/_{1 mole of oxygen}) = 24 x 10^{23} oxygen atoms
We can do the above math in our head by simply rounding 6.022 down to 6 and multiplying 6 by 4, then tacking on x 10^{23}. Rewriting our answer in proper scientific notation, we get 2.4 x 10^{24} oxygen atoms.

Question 2 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
2. Question
2. Which of the following samples contains the greatest number of moles of oxygen?
I. 54 g of H_{2}O
II. 98 g of H_{2}SO_{4}
III. 60 g of CH_{3}COOHCorrect / You marked this questionIn order to find how many moles of oxygen we have in each sample, we must first find how many moles of each sample we have. We can then use the molar ratio (moles of oxygen / moles of sample) to find the number of moles of oxygen in each sample. Sample I is 54 g of H_{2}O. Each mole of H_{2}O has a mass of 2(1g) + 16g = 18g. Thus, we have:
54 g of H_{2}O x (^{1 mol of H2O}/_{18 g of H}_{2}O) = 3 moles of H_{2}O
Each mole of H_{2}O contains one mole of oxygen, so we have 3 moles of oxygen in sample I. Sample II is 98 g of H_{2}SO_{4}. Each mole of H_{2}SO_{4} has a mass of 2(1g) + 32g + 4(16g) = 98g. Thus we have:
98 g of H_{2}SO_{4} x (^{1 mol of H2SO4}/_{98 g of H}_{2}SO_{4}) = 1 mole of H_{2}SO_{4}
Each mole of H_{2}SO_{4} contains 4 moles of oxygen, so we have 4 moles of oxygen in sample II. Sample III is 60 g of CH_{3}COOH. Each mole of CH_{3}COOH has a mass of 4(1g) + 2(12g) + 2(16g) = 60g. Thus we have:
60 g of CH_{3}COOH x (^{1 mol of CH3COOH}/_{60 g of CH}_{3}COOH) = 1 mole of CH_{3}COOH
Each mole of CH_{3}COOH contains 2 moles of oxygen, so we have 2 moles of oxygen in sample III. So, sample II has the most moles of oxygen.
Incorrect / You marked this questionIn order to find how many moles of oxygen we have in each sample, we must first find how many moles of each sample we have. We can then use the molar ratio (moles of oxygen / moles of sample) to find the number of moles of oxygen in each sample. Sample I is 54 g of H_{2}O. Each mole of H_{2}O has a mass of 2(1g) + 16g = 18g. Thus, we have:
54 g of H_{2}O x (^{1 mol of H2O}/_{18 g of H}_{2}O) = 3 moles of H_{2}O
Each mole of H_{2}O contains one mole of oxygen, so we have 3 moles of oxygen in sample I. Sample II is 98 g of H_{2}SO_{4}. Each mole of H_{2}SO_{4} has a mass of 2(1g) + 32g + 4(16g) = 98g. Thus we have:
98 g of H_{2}SO_{4} x (^{1 mol of H2SO4}/_{98 g of H}_{2}SO_{4}) = 1 mole of H_{2}SO_{4}
Each mole of H_{2}SO_{4} contains 4 moles of oxygen, so we have 4 moles of oxygen in sample II. Sample III is 60 g of CH_{3}COOH. Each mole of CH_{3}COOH has a mass of 4(1g) + 2(12g) + 2(16g) = 60g. Thus we have:
60 g of CH_{3}COOH x (^{1 mol of CH3COOH}/_{60 g of CH}_{3}COOH) = 1 mole of CH_{3}COOH
Each mole of CH_{3}COOH contains 2 moles of oxygen, so we have 2 moles of oxygen in sample III. So, sample II has the most moles of oxygen.

Question 3 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
3. Question
3. Below is the unbalanced chemical equation for the combustion of C_{2}H_{2}. From left to right, what are the coefficients in the balanced chemical equation?
__C_{2}H_{2} + __O_{2} → __CO_{2} + __H_{2}O
Correct / You marked this questionWhen balancing chemical equations, it is best to begin by balancing an element not present in more than one reactant or product. In this case, oxygen is present in both products, so we shouldn’t start by balancing oxygen. Let’s instead start by balancing the carbons. We have 2 carbons on the left in C_{2}H_{2}, but only one carbon on the right in CO_{2}. We balance the carbons by writing:
C_{2}H_{2} + O_{2} → 2CO_{2} + H_{2}O
The carbons are now balanced. This gives us 5 oxygens on the right side of the equation (2 x 2 = 4 from CO_{2} and 1 from H_{2}O). How can we get 5 oxygens on the left side of the equation? By multiplying O_{2} by ^{5}/_{2}, like so:
C_{2}H_{2} + ^{5}/_{2} O_{2} → 2CO_{2} + H_{2}O
This way, we now have ^{5}/_{2} x 2 = 5 oxygens on the left of the equation. However, all coefficients in balanced chemical equations must be integers; fractions are unacceptable. In order to get rid of the denominator in ^{5}/_{2}, we multiply the entire equation by 2, giving:
2C_{2}H_{2} + 5O_{2} → 4CO_{2} + 2H_{2}O
We now have 4 carbons on each side, 4 hydrogens on each side, and 10 oxygens on each side, meaning our equation is balanced.
Incorrect / You marked this questionWhen balancing chemical equations, it is best to begin by balancing an element not present in more than one reactant or product. In this case, oxygen is present in both products, so we shouldn’t start by balancing oxygen. Let’s instead start by balancing the carbons. We have 2 carbons on the left in C_{2}H_{2}, but only one carbon on the right in CO_{2}. We balance the carbons by writing:
C_{2}H_{2} + O_{2} → 2CO_{2} + H_{2}O
The carbons are now balanced. This gives us 5 oxygens on the right side of the equation (2 x 2 = 4 from CO_{2} and 1 from H_{2}O). How can we get 5 oxygens on the left side of the equation? By multiplying O_{2} by ^{5}/_{2}, like so:
C_{2}H_{2} + ^{5}/_{2} O_{2} → 2CO_{2} + H_{2}O
This way, we now have ^{5}/_{2} x 2 = 5 oxygens on the left of the equation. However, all coefficients in balanced chemical equations must be integers; fractions are unacceptable. In order to get rid of the denominator in ^{5}/_{2}, we multiply the entire equation by 2, giving:
2C_{2}H_{2} + 5O_{2} → 4CO_{2} + 2H_{2}O
We now have 4 carbons on each side, 4 hydrogens on each side, and 10 oxygens on each side, meaning our equation is balanced.

Question 4 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
4. Question
4. All of the following statements about the molar mass of H_{2}O are correct EXCEPT for one. Which one is the EXCEPTION?Correct / You marked this questionThe masses given for different elements in the periodic table give the mass (in amu’s) of a single atom of that element. For example, the 12.011 under C tells us that one atom of carbon weighs roughly 12.011 amu. By connection, one single molecule of H_{2}O has a mass of 18.02 amu. The relationship between amu’s, grams and moles is as follows:
If you have 6.022 x 10^{23} atoms/molecules, that group will have a mass in grams equal to the mass of a single atom/molecule in amu. For example, one single molecule of H_{2}O has a mass of 18.02 amu; if we instead had 6.022 x 10^{23} molecules of H_{2}O (1 mole of H_{2}O), that mole would have a mass of 18.02 grams. Thus we can see that [A] and [C] are true, but [B] is false.
If we were to rewrite [B] correctly, it would read: 18.02 amu of H_{2}O contains one molecule of H_{2}O.
[D] is true, as each mole of H_{2}O contains 1 mole of oxygen atoms. Thus one mole of H_{2}O would contain 6.022 x 10^{23} oxygen atoms. [E] is also true, as 18.02 g of H_{2}O (1 mole of H_{2}O) contains 2 moles of hydrogen atoms.
Incorrect / You marked this questionThe masses given for different elements in the periodic table give the mass (in amu’s) of a single atom of that element. For example, the 12.011 under C tells us that one atom of carbon weighs roughly 12.011 amu. By connection, one single molecule of H_{2}O has a mass of 18.02 amu. The relationship between amu’s, grams and moles is as follows:
If you have 6.022 x 10^{23} atoms/molecules, that group will have a mass in grams equal to the mass of a single atom/molecule in amu. For example, one single molecule of H_{2}O has a mass of 18.02 amu; if we instead had 6.022 x 10^{23} molecules of H_{2}O (1 mole of H_{2}O), that mole would have a mass of 18.02 grams. Thus we can see that [A] and [C] are true, but [B] is false.
If we were to rewrite [B] correctly, it would read: 18.02 amu of H_{2}O contains one molecule of H_{2}O.
[D] is true, as each mole of H_{2}O contains 1 mole of oxygen atoms. Thus one mole of H_{2}O would contain 6.022 x 10^{23} oxygen atoms. [E] is also true, as 18.02 g of H_{2}O (1 mole of H_{2}O) contains 2 moles of hydrogen atoms.

Question 5 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
5. Question
5. Which of the following compounds has the largest % composition (by mass) of carbon?
I. CO_{2} II. C_{6}H_{12}O_{6} III. C_{3}H_{8}Correct / You marked this questionPercent composition by mass of an element in a compound is determined by the following formula:
Here our element of interest is carbon. There are 12g of carbon in each mole of CO_{2}. CO_{2} has a molar mass of 12g + 2(16g) = 44g/mole. We calculate the percent composition by mass of carbon in CO_{2} by writing:
Doing this math without a calculator, we could say that 12 is a little more than 1/4th (25%) of 44 because we know that 12/48 = 0.25. Next, there are 6(12g) = 72g of carbon in each mole of C_{6}H_{12}O_{6} (better known as glucose). Glucose has a molar mass of 6(12g) + 12(1g) + 6(16g) = 180g/mole. We calculate the percent composition by mass of carbon in C_{6}H_{12}O_{6} by writing:
Doing this math without a calculator, we could say that 72 is less than half of 180 (which is 90), but more than a quarter of 180 (180/4 is 45). Thus, we know the percent composition of carbon in sample II is greater than in sample I. Lastly, there are 3(12g) = 36g of carbon in each mole of C_{3}H_{8}. C_{3}H_{8} has a molar mass of 3(12g) + 8(1g) = 44g/mole. We calculate the percent composition by mass of carbon in C_{3}H_{8} by writing:
Doing this math without a calculator, we could reduce 36/44 down to a more manageable fraction like 9/11. This is going to be just a little smaller than 9/10 (90%), so we can be confident it is significantly larger than the percent composition of carbon in samples I and II.
Incorrect / You marked this questionPercent composition by mass of an element in a compound is determined by the following formula:
Here our element of interest is carbon. There are 12g of carbon in each mole of CO_{2}. CO_{2} has a molar mass of 12g + 2(16g) = 44g/mole. We calculate the percent composition by mass of carbon in CO_{2} by writing:
Doing this math without a calculator, we could say that 12 is a little more than 1/4th (25%) of 44 because we know that 12/48 = 0.25. Next, there are 6(12g) = 72g of carbon in each mole of C_{6}H_{12}O_{6} (better known as glucose). Glucose has a molar mass of 6(12g) + 12(1g) + 6(16g) = 180g/mole. We calculate the percent composition by mass of carbon in C_{6}H_{12}O_{6} by writing:
Doing this math without a calculator, we could say that 72 is less than half of 180 (which is 90), but more than a quarter of 180 (180/4 is 45). Thus, we know the percent composition of carbon in sample II is greater than in sample I. Lastly, there are 3(12g) = 36g of carbon in each mole of C_{3}H_{8}. C_{3}H_{8} has a molar mass of 3(12g) + 8(1g) = 44g/mole. We calculate the percent composition by mass of carbon in C_{3}H_{8} by writing:
Doing this math without a calculator, we could reduce 36/44 down to a more manageable fraction like 9/11. This is going to be just a little smaller than 9/10 (90%), so we can be confident it is significantly larger than the percent composition of carbon in samples I and II.

Question 6 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
6. Question
6. Given that a certain compound’s empirical formula is NO_{2} and its molar mass is 92g, find the compound’s molecular formula.Correct / You marked this questionWhen given the empirical formula of a compound and that compound’s molar mass, we calculate the molecular formula as follows:
We then multiply each of the subscripts in the empirical formula by “X” to find the molecular formula. We are told the molar mass of the compound of interest is 92g and we can calculate the mass of the empirical formula as 14g + 2(16g) = 46g. Plugging these values into our formula gives:
^{92g}/_{46g} = 2
Lastly, we multiply each of the subscripts in our empirical formula by 2, giving N_{2}O_{4} as our final answer.
Incorrect / You marked this questionWhen given the empirical formula of a compound and that compound’s molar mass, we calculate the molecular formula as follows:
We then multiply each of the subscripts in the empirical formula by “X” to find the molecular formula. We are told the molar mass of the compound of interest is 92g and we can calculate the mass of the empirical formula as 14g + 2(16g) = 46g. Plugging these values into our formula gives:
^{92g}/_{46g} = 2
Lastly, we multiply each of the subscripts in our empirical formula by 2, giving N_{2}O_{4} as our final answer.

Question 7 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
7. Question
7. Given the following equation, how many grams of H_{2}O (18g/mole) will be formed from the reaction of 64g of O_{2} (32g/mole) with excess H_{2} (2g/mole)?
H_{2} + O_{2} → H_{2}O
Correct / You marked this questionRemember we can use the acronym BCC to help us find product and reactant amounts in problems like this. The “B” in BCC stands for “balance the equation.” Notice that the equation is unbalanced as written; there are two oxygens on the left and only one on the right. Thus we multiply the right side by two, giving:
H_{2} + O_{2} → 2H_{2}O
Is our equation now balanced? No! There are now 4 H’s on the right and only 2 on the left. Thus we must multiply the H_{2} on the left by two, giving our final balanced equation:
2H_{2} + O_{2} → 2H_{2}O
The first “C” stands for “convert to moles.” We want to convert the reactant and/or product amounts given in the problem to moles. We are told that we react 64g of O_{2} with excess H_{2}. We can’t convert “excess H_{2}” into moles, but we can convert 64g of O_{2} to moles by using the molar mass of O_{2}:
64g of O_{2} x (^{1 mol of O2}/_{32g of O}_{2}) = 2 moles of O_{2}
The second “C” stands for “coefficients,” meaning we use the coefficients in the balanced chemical equation to convert our products into reactants. Our chemical equation tells us that for each 1 mole of O_{2} we react, we create 2 moles of H_{2}O. Thus, mathematically, we say:
2 moles of O_{2} x (^{2 moles of H2O}/_{1 mole of O}_{2}) = 4 moles of H_{2}O
Lastly, we convert our moles of H_{2}O into grams, as the question wanted the number of grams of H_{2}O we would form.
4 moles of H_{2}O x (^{18g of H2O}/_{1 mole of H}_{2}O) = 72g of H_{2}O
Incorrect / You marked this questionRemember we can use the acronym BCC to help us find product and reactant amounts in problems like this. The “B” in BCC stands for “balance the equation.” Notice that the equation is unbalanced as written; there are two oxygens on the left and only one on the right. Thus we multiply the right side by two, giving:
H_{2} + O_{2} → 2H_{2}O
Is our equation now balanced? No! There are now 4 H’s on the right and only 2 on the left. Thus we must multiply the H_{2} on the left by two, giving our final balanced equation:
2H_{2} + O_{2} → 2H_{2}O
The first “C” stands for “convert to moles.” We want to convert the reactant and/or product amounts given in the problem to moles. We are told that we react 64g of O_{2} with excess H_{2}. We can’t convert “excess H_{2}” into moles, but we can convert 64g of O_{2} to moles by using the molar mass of O_{2}:
64g of O_{2} x (^{1 mol of O2}/_{32g of O}_{2}) = 2 moles of O_{2}
The second “C” stands for “coefficients,” meaning we use the coefficients in the balanced chemical equation to convert our products into reactants. Our chemical equation tells us that for each 1 mole of O_{2} we react, we create 2 moles of H_{2}O. Thus, mathematically, we say:
2 moles of O_{2} x (^{2 moles of H2O}/_{1 mole of O}_{2}) = 4 moles of H_{2}O
Lastly, we convert our moles of H_{2}O into grams, as the question wanted the number of grams of H_{2}O we would form.
4 moles of H_{2}O x (^{18g of H2O}/_{1 mole of H}_{2}O) = 72g of H_{2}O

Question 8 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
8. Question
8. How many moles of CO_{2} will be formed from the complete combustion of 88 g of C_{3}H_{8} (44 g/mole) with 160 g of O_{2} (32 g/mole)?Correct / You marked this questionWe first identify this problem as a limiting reagent problem by noticing that the question stem gives us amounts of both reactants and asks us to predict the amount of product made. We can use the BCPAA acronym to answer limiting reagent questions. The “B” stands for “balance the chemical equation,” so we write a chemical equation for the combustion of C_{3}H_{8}:
C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O
After balancing the equation, it looks like this:
C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O
The “C” stands for “convert to moles,” so we convert any given reactant amounts to moles by using their molar masses.
88g of C_{3}H_{8} x (^{1 mol of C3H8}/_{44g of C}_{3}H_{8}) = 2 moles of C_{3}H_{8}
160g of O_{2} x (^{1 mol of O2}/_{32g of O}_{2}) = 5 moles of O_{2}
The “P” stands for “pick one,” meaning we should pick one of our reactants to investigate further. Let’s arbitrarily pick O_{2}. The first “A” stands for “ask;” we ask, “If we were to use up all of the O_{2} we have, how much C_{3}H_{8} would we use?” Our balanced chemical formula tells us that we use 1 mole of C_{3}H_{8} for every 5 moles of O_{2}, so we write:
5 moles of O_{2} x (^{1 mol of C3H8}/_{5 moles of O}_{2}) = 1 mole of C_{3}H_{8}
In words, this equation says that if we were to use up all 5 moles of O_{2} that we have, we would use up 1 mole of C_{3}H_{8}. The second “A” also stands for “ask;” we ask, “Do we have at least 1 mole of C_{3}H_{8}?” Yes, we have more than 1 mole of C_{3}H_{8}; therefore C_{3}H_{8} is not the limiting reagent. Now that we know O_{2} is our limiting reagent, we can use the coefficients in our chemical equation to see how many moles of CO_{2} we will form:
5 moles of O_{2} x (^{3 moles of CO2}/_{5 moles of O}_{2}) = 3 moles of CO_{2}
Incorrect / You marked this questionWe first identify this problem as a limiting reagent problem by noticing that the question stem gives us amounts of both reactants and asks us to predict the amount of product made. We can use the BCPAA acronym to answer limiting reagent questions. The “B” stands for “balance the chemical equation,” so we write a chemical equation for the combustion of C_{3}H_{8}:
C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O
After balancing the equation, it looks like this:
C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O
The “C” stands for “convert to moles,” so we convert any given reactant amounts to moles by using their molar masses.
88g of C_{3}H_{8} x (^{1 mol of C3H8}/_{44g of C}_{3}H_{8}) = 2 moles of C_{3}H_{8}
160g of O_{2} x (^{1 mol of O2}/_{32g of O}_{2}) = 5 moles of O_{2}
The “P” stands for “pick one,” meaning we should pick one of our reactants to investigate further. Let’s arbitrarily pick O_{2}. The first “A” stands for “ask;” we ask, “If we were to use up all of the O_{2} we have, how much C_{3}H_{8} would we use?” Our balanced chemical formula tells us that we use 1 mole of C_{3}H_{8} for every 5 moles of O_{2}, so we write:
5 moles of O_{2} x (^{1 mol of C3H8}/_{5 moles of O}_{2}) = 1 mole of C_{3}H_{8}
In words, this equation says that if we were to use up all 5 moles of O_{2} that we have, we would use up 1 mole of C_{3}H_{8}. The second “A” also stands for “ask;” we ask, “Do we have at least 1 mole of C_{3}H_{8}?” Yes, we have more than 1 mole of C_{3}H_{8}; therefore C_{3}H_{8} is not the limiting reagent. Now that we know O_{2} is our limiting reagent, we can use the coefficients in our chemical equation to see how many moles of CO_{2} we will form:
5 moles of O_{2} x (^{3 moles of CO2}/_{5 moles of O}_{2}) = 3 moles of CO_{2}

Question 9 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
9. Question
9. How many moles of MgO will be formed when 72.9 g of Mg and 64 g of O_{2} are reacted together, according to the following unbalanced chemical equation?
Mg + O_{2} → MgO
Correct / You marked this questionWe first identify this problem as a limiting reagent problem by noticing that the question stem gives us amounts of both reactants and asks us to predict the amount of product made. We can use the “BCPAA!” acronym to answer limiting reagent questions. The “B” stands for “balance the chemical equation.” We notice there are 2 O’s on the left and only one on the right, so we multiply MgO by 2. This balances our O’s but unbalances our Mg’s, so we multiply our Mg on the left by 2, giving our balanced equation:
2Mg + O_{2} → 2MgO
The “C” stands for “convert to moles,” so we convert any given reactant amounts to moles by using their molar masses:
72.9g of Mg x (^{1 mol of Mg}/_{24.3g of Mg}) = 3 moles of Mg
64g of O_{2} x (^{1 mol of O2}/_{32g of O}_{2}) = 2 moles of O_{2}
The “P” stands for “pick one,” meaning we should pick one of our reactants to investigate further. Let’s arbitrarily pick O_{2}. The first “A” stands for “ask;” we ask, “If we were to use up all of the O_{2} we have, how much Mg would we use?” Our balanced chemical formula tells us that we use 2 mole of Mg for every 1 mole of O_{2}, so we write:
2 moles of O2 x (^{2 moles of Mg}/_{1 mole of O}_{2}) = 4 moles of Mg
In words, this equation says that if we were to use up all 2 moles of O_{2} that we have, we would use up 4 moles of Mg. The second “A” also stands for “ask;” we ask, “Do we have at least 4 moles of Mg?” No, we only have 3 moles of Mg; therefore Mg is the limiting reagent. Now that we know Mg is our limiting reagent, we can use the coefficients in our chemical equation to see how many moles of MgO we will form:
3 moles of Mg x (^{2 moles of MgO}/_{2 moles of Mg}) = 3 moles of MgO
Incorrect / You marked this questionWe first identify this problem as a limiting reagent problem by noticing that the question stem gives us amounts of both reactants and asks us to predict the amount of product made. We can use the “BCPAA!” acronym to answer limiting reagent questions. The “B” stands for “balance the chemical equation.” We notice there are 2 O’s on the left and only one on the right, so we multiply MgO by 2. This balances our O’s but unbalances our Mg’s, so we multiply our Mg on the left by 2, giving our balanced equation:
2Mg + O_{2} → 2MgO
The “C” stands for “convert to moles,” so we convert any given reactant amounts to moles by using their molar masses:
72.9g of Mg x (^{1 mol of Mg}/_{24.3g of Mg}) = 3 moles of Mg
64g of O_{2} x (^{1 mol of O2}/_{32g of O}_{2}) = 2 moles of O_{2}
The “P” stands for “pick one,” meaning we should pick one of our reactants to investigate further. Let’s arbitrarily pick O_{2}. The first “A” stands for “ask;” we ask, “If we were to use up all of the O_{2} we have, how much Mg would we use?” Our balanced chemical formula tells us that we use 2 mole of Mg for every 1 mole of O_{2}, so we write:
2 moles of O2 x (^{2 moles of Mg}/_{1 mole of O}_{2}) = 4 moles of Mg
In words, this equation says that if we were to use up all 2 moles of O_{2} that we have, we would use up 4 moles of Mg. The second “A” also stands for “ask;” we ask, “Do we have at least 4 moles of Mg?” No, we only have 3 moles of Mg; therefore Mg is the limiting reagent. Now that we know Mg is our limiting reagent, we can use the coefficients in our chemical equation to see how many moles of MgO we will form:
3 moles of Mg x (^{2 moles of MgO}/_{2 moles of Mg}) = 3 moles of MgO

Question 10 of 10Mike's Videos  Ch. 1 Quiz: Stoichiometry
10. Question
10. Calculate the percent yield of C_{2}H_{2(g)} given that 28 g of C_{2}H_{2(g)} was produced when 128 g of CaC_{2(s)} was reacted with excess water according to the following balanced reaction:
CaC_{2(s)} + 2H_{2}O_{(l)} → Ca(OH)_{2(aq)} + C_{2}H_{2(g)}
Correct / You marked this questionTo calculate percent yield of a reaction, we divide the mass of the actual yield by the mass of the theoretical yield. In this case, we know the mass of the actual yield (28g), but we need to find the mass of the theoretical yield.
To find the theoretical yield, we start with the mass of the limiting reagent and use mole ratios found in the balanced chemical equation to convert to mass of the desired product. In this case, we know that CaC_{2(s)} is the limiting reagent because the other reactant, H_{2}O, is said to be in excess. Our calculations should look like this:
Next, we divide our actual yield (28g) by the theoretical yield (52g) and multiply by 100% to get our final answer:
Because this math is difficult to do in your head and you will not have a calculator during the Survey of the Natural Sciences section, it is best to get good at rounding. For example, you may not know what 28/52 is off the top of your head, but you should recognize that 28 is just two more than half of 52. Thus, we know our answer should be a percentage slightly more than 50%. [C] is the only answer even close to 50% and therefore must be correct.
Incorrect / You marked this questionTo calculate percent yield of a reaction, we divide the mass of the actual yield by the mass of the theoretical yield. In this case, we know the mass of the actual yield (28g), but we need to find the mass of the theoretical yield.
To find the theoretical yield, we start with the mass of the limiting reagent and use mole ratios found in the balanced chemical equation to convert to mass of the desired product. In this case, we know that CaC_{2(s)} is the limiting reagent because the other reactant, H_{2}O, is said to be in excess. Our calculations should look like this:
Next, we divide our actual yield (28g) by the theoretical yield (52g) and multiply by 100% to get our final answer:
Because this math is difficult to do in your head and you will not have a calculator during the Survey of the Natural Sciences section, it is best to get good at rounding. For example, you may not know what 28/52 is off the top of your head, but you should recognize that 28 is just two more than half of 52. Thus, we know our answer should be a percentage slightly more than 50%. [C] is the only answer even close to 50% and therefore must be correct.
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