# What volume of carbon dioxide will be released during the interaction of 10.6 g of sodium carbonate

**What volume of carbon dioxide will be released during the interaction of 10.6 g of sodium carbonate and with a solution of hydrochloric acid?**

First, we write down the equation of reactions.

Na2CO3 + 2 HCl = 2NaCl + H2O + CO2.

First, we find the amount of sodium carbonate substance. We write down the solution.

n = m / M.

M (Na2CO3) = 23 × 2 + 12 + 3 × 16 = 106 g / mol.

n = 10.6 g / 106 g / mol = 0.1 mol.

Next, we find the amount of carbon dioxide substance according to the reaction equation.

0.1 mol Na2CO3 – x mol CO2.

1 mol Na2CO3 – 1 mol CO2.

Find the unknown value of x.

X = 0.1 mol × 1 ÷ 1 = 0.1 mol CO2.

Next, we find the volume of carbon dioxide.

n = V / Vm.

V = n × Vm = 0.1 mol × 22.4 L / mol = 2.24 L.

This means that the result is 2.24 liters of carbon dioxide.

Answer: V (CO2) = 2.24 l.