What volume of carbon monoxide IV is formed by the interaction of 12 g of lithium carbonate with 9.8 g of sulfuric acid?

The reaction of dissolving lithium carbonate in sulfuric acid is described by the following chemical reaction equation:

Li2CO3 + H2SO4 = 2Li2SO4 + CO2 + H2O;

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

Let’s calculate the available chemical amount of lithium carbonate.

M Li2CO3 = 7 x 2 + 12 + 16 x 3 = 74 grams / mol;

N Li2CO3 = 12/74 = 0.162 mol;

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

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

N H2SO4 = 9.8 / 98 = 0.1 mol;

Thus, when 0.1 mol of salt is dissolved, 0.1 mol of carbon dioxide is synthesized.

This will consume 0.1 mol of acid.

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

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

V CO2 = 0.1 x 22.4 = 2.24 liters;