# 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.

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