When 105 g of nitric acid and 318 g of sodium carbonate interact, carbon dioxide will be released. Find its volume.

Given:
m (HNO3) = 105 g
m (Na2CO3) = 318 g

To find:
V (CO2) -?

Decision:
1) 2HNO3 + Na2CO3 => 2NaNO3 + CO2 ↑ + H2O;
2) Mr (HNO3) = Ar (H) + Ar (N) + Ar (O) * 3 = 1 + 14 + 16 * 3 = 63 g / mol;
Mr (Na2CO3) = Ar (Na) * 2 + Ar (C) + Ar (O) * 3 = 23 * 2 + 12 + 16 * 3 = 106 g / mol;
3) n (HNO3) = m (HNO3) / Mr (HNO3) = 105/63 = 1.67 mol;
4) n (Na2CO3) = m (Na2CO3) / Mr (Na2CO3) = 318/106 = 3 mol;
5) n (CO2) = n (Na2CO3) = 3 mol;
6) V (CO2) = n (CO2) * Vm = 3 * 22.4 = 67.2 liters.

Answer: The CO2 volume is 67.2 liters.