How do stoichiometric ratios relate to molar volume of a gas?

Jan 11, 2014

They don’t.

Explanation:

The molar volume of a gas is a constant at a given temperature and pressure.

However, you can use the stoichiometric or molar ratios to calculate molar volumes.

EXAMPLE

$\text{MgCO"_3"(s)" + "H"_2"SO"_4"(aq)" → "MgSO"_4"(aq)" + "H"_2"O(l)" + "CO"_2"(g)}$

A mass of 24.0 g of $\text{Mg}$ produced 7.07 L of ${\text{CO}}_{2}$ at 25.0 °C and a pressure of 0.987 bar. What is the molar volume of ${\text{CO}}_{2}$ at STP (1 bar and 0 °C)?

Solution

(a) We must calculate the moles of ${\text{MgCO}}_{3}$ and then use the molar ratio to get the moles of ${\text{CO}}_{2}$

24.0 color(red)(cancel(color(black)("g MgCO"_3))) × (1 color(red)(cancel(color(black)("mol MgCO"_3))))/(84.31 color(red)(cancel(color(black)("g MgCO"_3)))) × (1 "mol CO"_2)/(1 color(red)(cancel(color(black)("mol MgCO"_3)))) = "0.2847 mol CO"_2

(b) Next, we use the combined gas laws to calculate the volume of the ${\text{CO}}_{2}$ at STP.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \frac{{P}_{1} {V}_{1}}{{T}_{1}} = \frac{{P}_{2} {V}_{2}}{{T}_{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this formula to give

V_2 = V_1 × (P_1)/(P_2) × (T_2)/(T_1)

${P}_{1} = \text{0.987 bar"; V_1 = "7.15 L"; T_1 = "298.15 K}$
${P}_{2} = \text{1 bar"; color(white)(mm)V_2 = ?;color(white)(mml) T_2 = "273.15 K}$

${V}_{2} = \text{7.15 L" × (0.987 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar")))) × (273.15 color(red)(cancel(color(black)("K"))))/(298.15 color(red)(cancel(color(black)("K")))) = "6.466 L}$

(c) Finally, we calculate the molar volume.

$\text{Molar volume" = "6.466 L"/"0.2847 mol" = "22.7 L/mol}$

The molar volume is 22.7 L at STP.