What volume of carbon dioxide is released when an excess of sodium carbonate solution 292 g

What volume of carbon dioxide is released when an excess of sodium carbonate solution 292 g interacts with hydrochloric acid solutions with a mass fraction of a solute of 10%

The reaction of sodium carbonate with hydrochloric acid is described by the following chemical reaction equation.

NaCO3 + 2HCl = 2NaCl + CO2;

When 1 mol of sodium carbonate is dissolved, 1 mol of carbon dioxide is released. This requires 2 mol of hydrochloric acid.

Determine the amount of hydrochloric acid substance contained in 292 grams of its 10% solution.

M HCl = 1 + 35.5 = 36.5 grams / mol;

m HCl = 292 x 0.1 = 29.2 grams;

N HCl = 29.2 / 36.5 = 0.8 mol;

Reaction with 0.8 mol of hydrochloric acid will produce 0.4 mol of carbon dioxide.

One mole of ideal gas under normal conditions takes a volume of 22.4 liters.

Let’s determine the volume of carbon dioxide.

V CO2 = 0.4 x 22.4 = 8.96 liters;