# Question a773b

Apr 14, 2015

The volume of gas is 3.571 L.

Step 1. Write the balanced equation.

${\text{CaCO"_3 → "CaO + CO}}_{2}$

Step 2. Calculate the moles of ${\text{CO}}_{2}$

${\text{Moles of CO"_2 = 15.74 cancel("g CaCO₃") × (1 cancel("mol CaCO₃"))/(100.09 cancel("g CaCO₃")) × ("1 mol CO"_2)/(1 cancel("mol CaCO₃")) = "0.157 26 mol CO}}_{2}$

Step 3. Use the Ideal Gas Law to calculate the volume of CO₂

$P V = n R T$

$V = \frac{n R T}{P}$

STP is 100 kPa and 273.15 K.

V = 0.157 26 cancel("mol") × (8.314 cancel("kPa")·"L" ·cancel("K⁻¹mol⁻¹") × 273.15 cancel("K"))/(100 cancel("kPa")) = "3.571 L"#

Apr 14, 2015

The great thing about STP is that the ideal gas law calculations have already been done for you for 1 mole of a gas.

At STP conditions, 100.0 kPa and 0 degrees C, to be more precise, 1 mole of any ideal gas occupies exactly 22.71 L. This is known as the molar volume of a gas at STP.

So, in your case, once you get the number of moles of carbon dioxide produced, you can multiply that value by the molar volume of a gas at STP to get the volume you need

$V = n \cdot {V}_{\text{molar" = 0.15726cancel("moles") * "22.71 L"/(1cancel("mol")) = "3.571 L}}$

So, don't forget about the molar volume of a gas at STP. If nothing else, you can use it to double check the calculations you've made using the ideal gas law.

At STP, for n moles of ideal gas

$P V = n R T \implies \frac{V}{n} = {\underbrace{\frac{R T}{P}}}_{\textcolor{b l u e}{\text{22.71 L}}}$