What volume of CO2 will be released when 15 g of Na2CO3 reacts with 10% impurities with excess HCI?

univerkov | September 24, 2021 | education

1. Let’s make a chemical equation, not forgetting to put down the coefficients:

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

2. It is known that 15g Na2CO3 is 100%; then we find 90% Na2CO3.

3. Find the mass of Na2CO3:

m (Na2CO3) = 15 * 90% = 13.5g.

4. Let’s find the amount of substance Na2CO3:

n (Na2CO3) = m / M = 13.5 / 106 = 0.127 mol.

5. The equation shows:

n (Na2CO3) = n (CO2).

6. Find the volume of carbon dioxide:

V (CO2) = n * Vm = 0.127 * 22.4 = 2.85l.

Answer: V (CO2) = 2.85l.

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