How to calculate the volume or mass of carbon dioxide gas absorbed by a calcium hydroxide solution?

Question

Carbon dioxide reacts with calcium hydroxide as shown by the equation: $$\ce{Ca(OH)2 (aq) + CO2 (g) -> CaCO3 (s) + H2O (l)}$$ A solution containing $\pu{0.1 mol}$ of calcium hydroxide reacts with excess carbon dioxide at $\pu{298 K}$ and $\pu{100 kPa}$. Which statement about the reaction is correct?
(a) It consumes $\pu{4.4 L}$ of carbon dioxide gas
(b) It produces $\pu{24.5 L}$ of liquid water
(c) It consumes $\pu{0.5 mol}$ of carbon dioxide
(d) It produces $\pu{10.0 g}$ of solid calcium carbonate

I have assumed that if there are $\pu{0.1 mol}$ of calcium hydroxide would there also be $\pu{0.1 mol}$ of carbon dioxide (since they both have a stoichiometric coefficient of 1?), please correct me if I am wrong.

Under this assumption, I have calculated the volume of carbon dioxide gas using the equation: $$n=\frac{V}{V_\mathrm{m}},$$ where $n = \pu{0.1 mol}$ and $V_\mathrm{m} = \pu{24.79 L/mol}$, as this is the volume one mole of particles will be contained in at a temperature of $\pu{298 K}$ and a pressure of $\pu{100 kPa}$.

Solving for $V$, I got: $$ V(\ce{CO2}) = n \cdot V_\mathrm{m} = \pu{0.1 mol} \cdot \pu{24.79 L//mol} = \pu{2.5 L} \text{(1 dp)}$$

However, from the statements above, $\pu{2.5 L}$ of carbon dioxide gas is not an option. I am not sure whether I have calculated the volume wrong or if I am supposed to use another approach for this question.

Answer 1

(a) It consumes $\pu{4.4 L}$ of carbon dioxide gas

(wrong)

You've correctly calculated the volume of carbon dioxide $V(\ce{CO2})$ via the molar volume of ideal gas $V_\mathrm{m} = \pu{24.790 L mol-3}$ at SATP (Standard Ambient Temperature and Pressure):

$$V(\ce{CO2}) = nV_\mathrm{m} = \pu{0.1 mol}\cdot\pu{24.790 L mol-3} \approx \pu{2.5 L}$$

I personally wouldn't recommend to remember molar volumes at STP, NTP, SATP etc. as the numbers are close, but still different, and it brings up a lot of confusion. Instead, only remember the ideal gas constant $R$, and remember it well as it's being used ubiquitously. Pretending you forgot the molar volume for these conditions, you can use ideal gas law directly:

$$V(\ce{CO2}) = \frac{nRT}{p} = \frac{\pu{0.1 mol}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}}{\pu{1e5 Pa}} \approx\pu{2.5 L}$$

Here, keep in mind that $\pu{1 Pa} = \pu{1 J m-3}$ and $\pu{1 m3} = \pu{1e3 L}$.

(b) It produces $\pu{24.5 L}$ of liquid water

(wrong)

Liquid water and SATP both imply that the density of water $ρ(\ce{H2O})\approx\pu{1 g mL-1}$, and its volume can be found via molecular mass $M(\ce{H2O})$:

$$V(\ce{H2O}) = \frac{m(\ce{H2O})}{ρ(\ce{H2O})} = \frac{nM(\ce{H2O})}{ρ(\ce{H2O})} = \frac{\pu{0.1 mol}\cdot\pu{18.02 g mol-1}}{\pu{1 g mL-1}} \approx \pu{1.8 mL}$$

(c) It consumes $\pu{0.5 mol}$ of carbon dioxide

(wrong)

The simplest out of four. As you've already ruled out, there are equimolar amounts of both reactants and products according to the stoichiometry, so $n(\ce{CO2}) = n(\ce{Ca(OH)2} = \pu{0.1 mol})$:

$$\ce{Ca(OH)2 (aq) + CO2 (g) -> CaCO3 (s) + H2O (l)}$$

(d) It produces $\pu{10.0 g}$ of solid calcium carbonate

(correct)

Mass of calcium carbonate $m(\ce{CaCO3})$ can be found trivially from its molar mass $M(\ce{CaCO3})$:

$$m(\ce{CaCO3}) = nM(\ce{CaCO3}) = \pu{0.1 mol}\cdot\pu{100.09 g mol-1} \approx \pu{10.0 g}$$