Q: If you start with 10.0 g of substance A and want to find the mass of substance B produced, which equation would you use first according to the diagram?

$$\text{moles A} = \frac{\text{mass A}}{\text{molar mass of A}}$$

Q: Suppose you have 5.00 g of reactant A (molar mass = 50.0 g/mol) and the reaction produces product B with a 2:1 mole ratio (A:B). What is the number of moles of B formed?

$$0.0500$$ moles \times \frac{1}{2} = 0.0250$ moles

Q: A reaction has the following stoichiometry: $$2A + 3B \rightarrow 4C. If $$you start with 10.0 g of A (molar mass = 20.0 g/mol) and excess B, what is the mass of C produced (molar mass = 25.0 g/mol)?

$$10.0 \text{ g A} \times \frac{1 \text{ mol A}}{20.0 \text{ g A}} \times \frac{4 \text{ mol C}}{2 \text{ mol A}} \times 25.0 \text{ g C} = 25.0 \text{ g C}$$

Q: In a reaction, 15.0 g of reactant A (molar mass = 30.0 g/mol) reacts to form product B (molar mass = 60.0 g/mol) with a 1:1 mole ratio. What is the mass of B formed?

$$15.0 \text{ g A} \times \frac{1 \text{ mol A}}{30.0 \text{ g A}} \times \frac{1 \text{ mol B}}{1 \text{ mol A}} \times 60.0 \text{ g B} = 30.0 \text{ g B}$$

Q: If you have 8.00 g of A (molar mass = 16.0 g/mol) and the reaction produces B with a 1:2 mole ratio (A:B), what is the mass of B formed if the molar mass of B is 10.0 g/mol?

$$8.00 \text{ g A} \times \frac{1 \text{ mol A}}{16.0 \text{ g A}} \times \frac{2 \text{ mol B}}{1 \text{ mol A}} \times 10.0 \text{ g B} = 10.0 \text{ g B}$$

Question 1

The diagram above represents a common sequence in stoichiometric calculations. Which of the following best describes the correct sequence of steps to convert the mass of substance A to the mass of substance B in a chemical reaction?

Accepted Answer

Convert mass A to moles A, use the mole ratio to find moles B, then convert moles B to mass B.

Question 2

If you start with 10.0 g of substance A and want to find the mass of substance B produced, which equation would you use first according to the diagram?

Accepted Answer

$$	ext{moles A} = \frac{	ext{mass A}}{	ext{molar mass of A}}$$

Question 3

In the context of the diagram, what is the purpose of the 'Amount A (in moles) to Amount B (in moles)' step?

Accepted Answer

To use the stoichiometric coefficients from the balanced chemical equation to relate moles of A to moles of B.

Question 4

Suppose you have 5.00 g of reactant A (molar mass = 50.0 g/mol) and the reaction produces product B with a 2:1 mole ratio (A:B). What is the number of moles of B formed?

Accepted Answer

$$0.0500$$ moles 	imes \frac{1}{2} = 0.0250$ moles

Question 5

Which of the following best explains why you cannot convert directly from mass of A to mass of B without using moles?

Accepted Answer

Because the stoichiometric relationship between A and B is based on moles, not mass.

Question 6

A reaction has the following stoichiometry: $$2A + 3B \rightarrow 4C. If $$you start with 10.0 g of A (molar mass = 20.0 g/mol) and excess B, what is the mass of C produced (molar mass = 25.0 g/mol)?

Accepted Answer

$$10.0 	ext{ g A} 	imes \frac{1 	ext{ mol A}}{20.0 	ext{ g A}} 	imes \frac{4 	ext{ mol C}}{2 	ext{ mol A}} 	imes 25.0 	ext{ g C} = 25.0 	ext{ g C}$$

Question 7

In a reaction, 15.0 g of reactant A (molar mass = 30.0 g/mol) reacts to form product B (molar mass = 60.0 g/mol) with a 1:1 mole ratio. What is the mass of B formed?

Accepted Answer

$$15.0 	ext{ g A} 	imes \frac{1 	ext{ mol A}}{30.0 	ext{ g A}} 	imes \frac{1 	ext{ mol B}}{1 	ext{ mol A}} 	imes 60.0 	ext{ g B} = 30.0 	ext{ g B}$$

Question 8

Why is it necessary to use the molar mass of both A and B in the conversion from mass of A to mass of B?

Accepted Answer

Because you must convert mass to moles for both substances to use the stoichiometric ratio.

Question 9

If you have 8.00 g of A (molar mass = 16.0 g/mol) and the reaction produces B with a 1:2 mole ratio (A:B), what is the mass of B formed if the molar mass of B is 10.0 g/mol?

Accepted Answer

$$8.00 	ext{ g A} 	imes \frac{1 	ext{ mol A}}{16.0 	ext{ g A}} 	imes \frac{2 	ext{ mol B}}{1 	ext{ mol A}} 	imes 10.0 	ext{ g B} = 10.0 	ext{ g B}$$

Question 10

Which step in the diagram would require you to use the coefficients from the balanced chemical equation?

Accepted Answer

The conversion from Amount A (in moles) to Amount B (in moles).

lecture 6 (ns) Stoichiometry, Limiting Reactants, and Percent Yield: Study Notes

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