Q: A container holds 200 screws, each with an average mass of 3.63 g. What is the total mass of screws in the container?

$$200 \times 3.63\ \mathrm{g} = 726\ \mathrm{g}$$

Q: How many atoms are present in 2.45 mol of copper ($$\mathrm{Cu}$$)?

$$2.45\ \mathrm{mol} \times 6.022 \times 10$$^{23}$$\ \mathrm{atoms/mol} = 1.48 \times 10$$^{24}$$\ \mathrm{atoms}$$

Q: What is the molar mass of water ($$\mathrm{H_2O}$$)?

$$2 \times 1.01\ \mathrm{g/mol} + 16.00\ \mathrm{g/mol} = 18.02\ \mathrm{g/mol}$$

Q: A student mixes hydrochloric acid ($$\mathrm{HCl}) $$with sodium bicarbonate ($$\mathrm{$$NaHCO_3$$}) $$and collects 2.50 g of $$\mathrm{CO_2$}$$ gas. If the theoretical yield is 3.00 g, what is the percent yield?

$$\frac{2.50\ \mathrm{g}}{3.00\ \mathrm{g}} \times 100\% = 83.3\%$$

Q: In the combustion of octane ($$2\ \mathrm{C_8H_{18}}(l) + 25\ \mathrm{O_2$}(g) \rightarrow 16\ \mathrm{CO_2$}(g) + 18\ \mathrm{H_2O}(g)$$), how many grams of $$\mathrm{CO_2$}$$ can be produced from 59.3 g of $$\mathrm{C_8H_{18}}$$?

$$59.3\ \mathrm{g}\ \mathrm{C_8H_{18}} \times \frac{1\ \mathrm{mol}}{114.26\ \mathrm{g}} \times \frac{16\ \mathrm{mol}\ \mathrm{CO_2$}}{2\ \mathrm{mol}\ \mathrm{C_8H_{18}}} \times \frac{44.01\ \mathrm{g}}{1\ \mathrm{mol}} = 183\ \mathrm{g}\ \mathrm{CO_2$}$$

Q: A reaction has a theoretical yield of 5.31 g $$\mathrm{N_2$}$$, but only 4.29 g is actually produced. What is the percent yield?

$$\frac{4.29\ \mathrm{g}}{5.31\ \mathrm{g}} \times 100\% = 80.8\%$$

Question 1

Which of the following statements best explains why chemists use the mole as a counting unit for atoms and molecules?

Accepted Answer

Atoms and molecules are too small to count individually, so the mole allows chemists to relate mass to number of particles using Avogadro's number: $$6.022 	imes 10$$^{23}$$.

Question 2

A container holds 200 screws, each with an average mass of 3.63 g. What is the total mass of screws in the container?

Accepted Answer

$$200 	imes 3.63\ \mathrm{g} = 726\ \mathrm{g}$$

Question 3

How many atoms are present in 2.45 mol of copper ($$\mathrm{Cu}$$)?

Accepted Answer

$$2.45\ \mathrm{mol} 	imes 6.022 	imes 10$$^{23}$$\ \mathrm{atoms/mol} = 1.48 	imes 10$$^{24}$$\ \mathrm{atoms}$$

Question 4

Which has more atoms: 5 g of hydrogen atoms or 5 g of lead atoms?

Accepted Answer

5 g of hydrogen atoms, because hydrogen has a much smaller molar mass ($$1.01\ \mathrm{g/mol}) $$than lead ($$207.2\ \mathrm{g/mol}$$), resulting in more moles and thus more atoms.

Question 5

What is the molar mass of water ($$\mathrm{H_2O}$$)?

Accepted Answer

$$2 	imes 1.01\ \mathrm{g/mol} + 16.00\ \mathrm{g/mol} = 18.02\ \mathrm{g/mol}$$

Question 6

Given the balanced equation $$2\ \mathrm{H_2$}(g) + \mathrm{O_2$}(g) \rightarrow 2\ \mathrm{H_2O}(l)$$, how many moles of water are produced from 5 moles of $$\mathrm{O_2$}$$?

Accepted Answer

$$2\ \mathrm{mol}\ \mathrm{H_2O} / 1\ \mathrm{mol}\ \mathrm{O_2$} 	imes 5\ \mathrm{mol}\ \mathrm{O_2$} = 10\ \mathrm{mol}\ \mathrm{H_2O}$$

Question 7

A student mixes hydrochloric acid ($$\mathrm{HCl}) $$with sodium bicarbonate ($$\mathrm{$$NaHCO_3$$}) $$and collects 2.50 g of $$\mathrm{CO_2$}$$ gas. If the theoretical yield is 3.00 g, what is the percent yield?

Accepted Answer

$$\frac{2.50\ \mathrm{g}}{3.00\ \mathrm{g}} 	imes 100\% = 83.3\%$$

Question 8

Which of the following best describes the limiting reactant in a chemical reaction?

Accepted Answer

The reactant that is completely consumed first, thus limiting the amount of product formed.

Question 9

In the combustion of octane ($$2\ \mathrm{C_8H_{18}}(l) + 25\ \mathrm{O_2$}(g) \rightarrow 16\ \mathrm{CO_2$}(g) + 18\ \mathrm{H_2O}(g)$$), how many grams of $$\mathrm{CO_2$}$$ can be produced from 59.3 g of $$\mathrm{C_8H_{18}}$$?

Accepted Answer

$$59.3\ \mathrm{g}\ \mathrm{C_8H_{18}} 	imes \frac{1\ \mathrm{mol}}{114.26\ \mathrm{g}} 	imes \frac{16\ \mathrm{mol}\ \mathrm{CO_2$}}{2\ \mathrm{mol}\ \mathrm{C_8H_{18}}} 	imes \frac{44.01\ \mathrm{g}}{1\ \mathrm{mol}} = 183\ \mathrm{g}\ \mathrm{CO_2$}$$

Question 10

A reaction has a theoretical yield of 5.31 g $$\mathrm{N_2$}$$, but only 4.29 g is actually produced. What is the percent yield?

Accepted Answer

$$\frac{4.29\ \mathrm{g}}{5.31\ \mathrm{g}} 	imes 100\% = 80.8\%$$

Question 11

Which of the following is the correct balanced equation for the combustion of methane?

Accepted Answer

$$\mathrm{CH_4$}(g) + 2\ \mathrm{O_2$}(g) \rightarrow \mathrm{CO_2$}(g) + 2\ \mathrm{H_2O}(g)$$

Question 12

If you have 8 molecules of $$\mathrm{O_2$}$$ and 5 molecules of $$\mathrm{CH_4$}$$, according to the reaction $$\mathrm{CH_4$} + 2\ \mathrm{O_2$} \rightarrow \mathrm{CO_2$} + 2\ \mathrm{H_2O}$$, what is the maximum number of $$\mathrm{CO_2$}$$ molecules that can be formed?

Accepted Answer

4 molecules, because $$\mathrm{O_2$} is $$the limiting reactant.

Question 13

Which of the following is a correct statement about the Law of Conservation of Mass as it applies to chemical reactions?

Accepted Answer

The total mass of reactants equals the total mass of products; atoms are neither created nor destroyed.

Question 14

How many molecules are in 453 g of $$\mathrm{CO_2$}$$? ($$\mathrm{Molar\ mass\ of\ CO_2$} = 44.01\ \mathrm{g/mol})$$

Accepted Answer

$$453\ \mathrm{g} 	imes \frac{1\ \mathrm{mol}}{44.01\ \mathrm{g}} 	imes 6.022 	imes 10$$^{23}$$\ \mathrm{molecules/mol} = 6.20 	imes 10$$^{24}$$\ \mathrm{molecules}$$

Question 15

A student needs to prepare 0.0265 g of pure carbon from pencil lead. How many carbon atoms does this sample contain? ($$\mathrm{Molar\ mass\ of\ C} = 12.01\ \mathrm{g/mol})$$

Accepted Answer

$$0.0265\ \mathrm{g} 	imes \frac{1\ \mathrm{mol}}{12.01\ \mathrm{g}} 	imes 6.022 	imes 10$$^{23}$$\ \mathrm{atoms/mol} = 1.33 	imes 10$$^{21}$$\ \mathrm{atoms}$$

Question 16

Which of the following is the correct molar mass for chlorine gas ($$\mathrm{Cl_2$}$$)?

Accepted Answer

$$2 	imes 35.45\ \mathrm{g/mol} = 70.90\ \mathrm{g/mol}$$

Question 17

If 87.4 g of $$\mathrm{NaOH}$$ reacts with excess $$\mathrm{Cl_2$}$$, how many grams of $$\mathrm{Cl_2$}$$ are needed? (Balanced equation: $$2\ \mathrm{NaOH} + \mathrm{Cl_2$} \rightarrow \mathrm{NaOCl} + \mathrm{NaCl} + \mathrm{H_2O}$$; $$\mathrm{Molar\ mass\ NaOH} = 40.00\ \mathrm{g/mol}$$, $$\mathrm{Molar\ mass\ Cl_2$} = 70.90\ \mathrm{g/mol})$$

Accepted Answer

$$87.4\ \mathrm{g}\ \mathrm{NaOH} 	imes \frac{1\ \mathrm{mol}}{40.00\ \mathrm{g}} 	imes \frac{1\ \mathrm{mol}\ \mathrm{Cl_2$}}{2\ \mathrm{mol}\ \mathrm{NaOH}} 	imes 70.90\ \mathrm{g} = 77.46\ \mathrm{g}\ \mathrm{Cl_2$}$$

Stoichiometry and Mass Relationships in Chemical Reactions

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