What volume of carbon dioxide (n.u.) is formed by the interaction of 5.3 g of sodium carbonate with 22 ml

What volume of carbon dioxide (n.u.) is formed by the interaction of 5.3 g of sodium carbonate with 22 ml of 25% sulfuric acid (density 1.78 g / ml)? What salt was formed in the solution?

The reaction of sodium carbonate dissolution in sulfuric acid is described by the following chemical reaction equation:

Na2CO3 + H2SO4 = Na2SO4 + CO2 + H2O;

When 1 mol of sodium carbonate is dissolved in acid, 1 mol of gaseous carbon dioxide is synthesized. This consumes 1 mol of sulfuric acid. This synthesizes sodium sulfate.

Let’s calculate the available chemical amount of soda.

M Na2CO3 = 23 x 2 + 12 + 16 x 3 = 106 grams / mol;

N Na2CO3 = 5.3 / 106 = 0.05 mol;

Let’s calculate the available chemical amount of sulfuric acid.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;

N H2SO4 = 22 x 1.78 x 0.25 / 98 = 0.1 mol;

Thus, when 0.05 mol of soda is dissolved, 0.05 mol of carbon dioxide is synthesized.

This will consume 0.05 mol of acid.

Let’s calculate its volume. To do this, multiply the amount of the substance and the standard volume of 1 mole of the gaseous substance.

1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

V CO2 = 0.05 x 22.4 = 1.12 liters;