When 63 g of hydrochloric acid reacted with 11.2 g of sodium carbonate, a gas was formed. find the volume of gas.

To solve, we compose the equation of the process:

2HCl + Na2CO3 = 2NaCl + CO2 + H2O – ion exchange, carbon dioxide is released;

Calculations according to the formulas of substances:
M (HCl) = 36.5 g / mol;

M (Na2CO3) = 105.8 g / mol;

M (CO2) = 44 g / mol.

Let’s determine the amount of starting materials:
Y (HCl) = m / M = 63 / 36.5 = 1.72 mol (substance in excess);

Y (Na2CO3) = m / M = 11.2 / 105.8 = 0.1 mol (deficient substance);

Y (CO2) = 0.1 mol since the amount of substances is 1 mol.

Calculations are made for the substance in deficiency.

Find the volume of the product:
V (CO2) = 0.1 * 22.4 = 2.24 l

Answer: the volume of carbon monoxide (4) is 2.24 liters