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
"MgCO"_3"(s)" + "H"_2"SO"_4"(aq)" → "MgSO"_4"(aq)" + "H"_2"O(l)" + "CO"_2"(g)"MgCO3(s)+H2SO4(aq)→MgSO4(aq)+H2O(l)+CO2(g)
A mass of 24.0 g of "Mg"Mg produced 7.07 L of "CO"_2CO2 at 25.0 °C and a pressure of 0.987 bar. What is the molar volume of "CO"_2CO2 at STP (1 bar and 0 °C)?
Solution
(a) We must calculate the moles of "MgCO"_3MgCO3 and then use the molar ratio to get the moles of "CO"_2CO2
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 "CO"_2 at STP.
color(blue)(bar(ul(|color(white)(a/a)(P_1V_1)/(T_1) = (P_2V_2)/(T_2)color(white)(a/a)|)))" "
We can rearrange this formula to give
V_2 = V_1 × (P_1)/(P_2) × (T_2)/(T_1)
P_1 = "0.987 bar"; V_1 = "7.15 L"; T_1 = "298.15 K"
P_2 = "1 bar"; color(white)(mm)V_2 = ?;color(white)(mml) T_2 = "273.15 K"
∴ V_2 = "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.
"Molar volume" = "6.466 L"/"0.2847 mol" = "22.7 L/mol"
The molar volume is 22.7 L at STP.