# Find the mass of carbon monoxide (IV) obtained by the action of hydrochloric acid on sodium carbonate

**Find the mass of carbon monoxide (IV) obtained by the action of hydrochloric acid on sodium carbonate weighing 110 g, containing impurities by mass fraction of 2%.**

Let’s find the mass of Na2CO3 without impurities.

100% – 2% = 98%.

110 g – 100%,

x – 98%,

x = (110 g × 98%): 100% = 107.8 g.

Let’s find the amount of the substance Na2CO3.

n = m: M.

M (Na2CO3) = 106 g / mol.

n = 107.8 g: 106 g / mol = 1.02 mol.

Let’s find the quantitative ratios of substances.

Na2CO3 + 2HCl = 2NaCl + CO2 ↑ + H2O

There is 1 mol of CO2 for 1 mole of Na2CO3.

The substances are in quantitative ratios of 1: 1.

The amount of substance will be equal.

n (Na2CO3) = n (CO2) = 1.02 mol.

Let’s find the mass of CO2.

m = n × M.

M (CO2) = 44 g / mol.

m = 44 g / mol × 1.02 mol = 44.88 g.

Answer: 44.88 g.