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)"#

A mass of 24.0 g of #"Mg"# produced 7.07 L of #"CO"_2# at 25.0 °C and a pressure of 0.987 bar. What is the molar volume of #"CO"_2# at STP (1 bar and 0 °C)?

**Solution**

**(a)** We must calculate the moles of #"MgCO"_3# and then use the molar ratio to get the moles of #"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 #"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.