Home / AP Chemistry: 1.1 Moles and Molar Mass – Exam Style questions with Answer- MCQ

# AP Chemistry: 1.1 Moles and Molar Mass – Exam Style questions with Answer- MCQ

Questions

Which of the following is the net ionic equation for the overall reaction that occurs as the cell operates?

(A) $$Ag(s) + Zn^{2+}(aq) \rightarrow Ag^{2+}(aq) + Zn(s)$$
(B)  $$Ag^{+}(aq) + Zn^{2+}(aq) + 3 e \rightarrow AgZn(s)$$
(C) $$Ag^{+}(aq) + Zn(s) \rightarrow Ag(s) + Zn^{2+}$$(aq) + e^{-}\)
(D) $$2 Ag^{+}(aq) + Zn(s) \rightarrow 2 Ag(s) + Zn^{2+}$$(aq)

Ans: D

To determine the net ionic equation for the overall reaction, we need to consider the two half-reactions and their standard reduction potentials given in the table:

Agt(aq) + e -+ Ag(s) E° = +0.80 V Zn2+(aq) + 2 e- – Zn(s) E° = -0.76 V

The half-reaction with the more positive reduction potential (Ag/Ag) will be the reduction reaction, while the half-reaction with the less positive (or more negative) reduction potential (Zn2+/Zn) will be the
oxidation reaction.

The overall reaction is obtained by multiplying the two half-reactions to cancel out the electrons transferred and then adding them:

Reduction: Ag*(aq) + e — > Ag(s) Oxidation: Zn(s) -> Zn2+(ag) + 2 e-

Overall reaction: Agt(aq) + Zn(s) -> Ag(s) + Zn2+(aq)

Therefore, the correct answer is (C): Ag+(aq) + Zn(s) -> Ag(s) + Zn2+(aq) + e-

Questions

Another sample of eggshell reacts completely with 4.0 mL of an HCl(aq) solution of unknown concentration. If the reaction produced 0.095 atm of gas, the concentration of the HCl(aq) solution was at least
(A) 0.0020 M
(B) 0.050 M
(C) 0.50 M
(D) 1.0 M

Ans: D

$$P=0.095$$ atm $$\rightarrow$$ corresponds to $$0.20 \mathrm{~g} \mathrm{CaCO}_3$$
So, moles of $$\mathrm{CaCO}_3=\frac{0.20 \mathrm{~g}}{100 \mathrm{gm}{mol}^{-1}}=0.002 \mathrm{moles}$$
moles of $$\mathrm{HCl}=\left(\right.$$ moles of $$\left.\mathrm{CaCO}_3 \times 2\right) \Rightarrow 0.002 \times 2$$
So, molarity of $$\mathrm{HCl}=\frac{\text { moles }}{\text { Volume in } \mathrm{L}} \Rightarrow \frac{0.004}{4} \times 1000$$
Concentration of $$\mathrm{HCl}=1 \mathrm{~m} \rightarrow$$ option $$(\mathrm{D})$$

Questions

The mass percent of $$CaCO_{3}$$(s) in the eggshell  is closest to
(A) 30%
(B) 45%
(C) 60%
(D) 75%

Ans: D

(36) Pressure of gas $$=0.870-0.800=0.07 \mathrm{~atm}$$ Mass of $$\mathrm{CaCO}_3$$ corresponding to it $$\Rightarrow 0.15 \mathrm{~g}$$ (approx)
So, $$\% \mathrm{CaCO}_3=\frac{0.15}{0.20} \times 100 \Rightarrow 75 \%$$

Questions

$$CO_2(g) + 2 LiOH(s) → Li_2CO3(aq) + H_2O$$(l)

In a one-person spacecraft, an astronaut exhales 880 g of $$CO_2$$(g) (molar mass 44 g/mol) per day. To prevent the buildup of $$CO_{2}$$(g) in the spacecraft, a device containing LiOH(s) is used to remove the $$CO_2$$(g), as represented by the equation above. What mass of LiOH(s) (molar mass 24 g/mol) is needed to react with all of the $$CO_2$$(g) produced by an astronaut in one day?
(A) 40. g
(B) 240 g
(C) 480 g
(D) 960 g

Ans: D

First, let’s calculate the moles of $$CO_2$$ produced by the astronaut in one day:

Mass of$$CO_2$$ produced per day = 880 g
Molar mass of $$CO_2$$ =44 g/mol

$\text{moles of } CO_2 =\frac{\text{mass of } CO_2}{\text{molar mass of } CO_2}$

$\text{moles of } CO_2 = \frac{880 \, \text{g}}{44 \, \text{g/mol}} = 20 \, \text{mol}$

According to the balanced chemical equation, 1 mole of $$CO_2$$ reacts with 2 moles of $$LiQH$$. So, to find the moles of $$LiQH$$ needed, we multiply the moles of $$CO_2$$ by the stoichiometric coefficient ratio:

$\text{moles of } LiQH = 2 \times \text{moles of } CO_2$

$\text{moles of } LiQH = 2 |times 20 , \text{mol} = 40 \, \text{mol}$

Now, let’s calculate the mass of $$LiQH$$ needed:

Molar mass of $$LiQH$$ = 24 g/mol

$\text{mass of } LiQH = \text{moles of } LiQH \times \text{molar mass of } LiQH$

$\text{mass of } LiQH = 40 \, \text{mol} \times 24 \, \text{g/mol} = 960 \, \text{g}$

So, the mass of $$LiQH$$ needed to react with all of the $$CO_2$$ produced by the astronaut in one day is 960 g.

Therefore, the correct answer is (D) 960 g.

### Question

$\mathrm{Li}_3 \mathrm{~N}(s)+2 \mathrm{H}_2(g) \rightleftarrows \mathrm{LiNH}_2(s)+2 \mathrm{LiH}(s) \quad \Delta H^{\circ}=-192 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

Because pure $$\mathrm{H}_2$$ is a hazardous substance, safer and more cost effective techniques to store it as a solid for shipping purposes have been developed. One such method is the reaction represented above, which occurs at $$200^{\circ} \mathrm{C}$$.

When 70. g of$$Li_3$$N(s) (molar mass 35 g/mol)  reacts with excess H2(g), 8.0 g of $$LiH(s)$$ is produced. The percent yield is closest to
(A) 17%
(B) 25%
(C) 50.%
(D) 100%

Ans:B

To find the percent yield of $$\mathrm{LiH}(s)$$, we first need to determine the theoretical yield of $$\mathrm{LiH}(s)$$ from the given reaction and then compare it to the actual yield.

1. Calculate the number of moles of $$\mathrm{Li}_3\mathrm{N}(s)$$ used:
$\text{Moles of } \mathrm{Li}_3\mathrm{N}(s) = \frac{\text{mass}}{\text{molar mass}} = \frac{70 \, \text{g}}{35 \, \text{g/mol}} = 2 \, \text{mol}$

2. Using the stoichiometry of the reaction, determine the number of moles of $$\mathrm{LiH}(s)$$ produced:
From the balanced equation, we see that the coefficient of $$\mathrm{LiH}(s)$$ is 2. This means that for every 1 mole of $$\mathrm{Li}_3\mathrm{N}(s)$$ that reacts, 2 moles of $$\mathrm{LiH}(s)$$ are produced.

So, moles of $$\mathrm{LiH}(s)$$ produced = $$2 \times \text{moles of } \mathrm{Li}_3\mathrm{N}(s) = 2 \times 2 \, \text{mol} = 4 \, \text{mol}$$

3. Calculate the theoretical yield of $$\mathrm{LiH}(s)$$:
$\text{Theoretical yield} = \text{moles of } \mathrm{LiH}(s) \times \text{molar mass of } \mathrm{LiH}(s)$
$\text{Theoretical yield} = 4 \, \text{mol} \times \frac{6.94 + 1.008}{1} \, \text{g/mol} = 4 \times 7.948 = 31.792 \, \text{g}$

4. Calculate the percent yield:
$\text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\%$
$\text{Percent yield} = \left( \frac{8.0 \, \text{g}}{31.792 \, \text{g}} \right) \times 100\% \approx 25\%$

So, the closest answer choice is: (B) 25%

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