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

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.

• OK, you eliminated (a). How about (b), (c) and (d)? – MaxW Feb 12 '19 at 8:05
• I know that it can't be (c) either, since the reaction consumes 0.1 moles of carbon dioxide, not 0.5. Just to clarify, would the number of moles of CaCO3 and H2O also be 0.1 (since all the reactants and products have a stoichiometric coefficient of 1)? – ceno980 Feb 12 '19 at 9:35
• Correct (c) is wrong too. // Yes, the coefficients in balanced equation give the molar ratios. So 0.1 mole of calcium hydroxide and 0.1 mole of Carbon dioxide yield 0.1 mole of calcium carbonate and 0.1 mole of water. – MaxW Feb 12 '19 at 9:40
• Okay, I got the answer, it's (d). Thank you for your help. – ceno980 Feb 12 '19 at 9:54
• OP (or @MaxW) I think the question is quite interesting for a homework problem. Maybe one of you could provide an answer with the rest of the way to the solution. (Self-answers are encouraged.) – Martin - マーチン Feb 12 '19 at 13:54

(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 $$M(\ce{CaCO3})$$:

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