When carbon dioxide was passed through a solution of sodium hydroxide with a volume

When carbon dioxide was passed through a solution of sodium hydroxide with a volume of 4.9 liters, a mixture of carbonate and sodium bicarbonate weighing 22.9 g was obtained. Calculate the composition of the resulting mixture of salts

1. Let’s compose the reaction equations:

2NaOH + CO2 = Na2CO3 + H2O;

NaOH + CO2 = NaHCO3;

2. let the chemical amount of sodium carbonate x mol, then the amount spent on its production of carbon dioxide x mol;

3. The amount of bicarbonate is taken as a mole, and, accordingly, the amount of carbon dioxide in this equation is also in a mole;

4.Calculate the chemical amount of carbon dioxide:

n (CO2) = V (CO2): Vm = 4.9: 22.4 = 0.21875 mol;

n (CO2) = x + y = 0.21875 mol;

5.we express the masses of the formed salts:

m (Na2CO3) = 106x;

m (NaHCO3) = 84y;

6. Let’s compose a system of equations:

x + y = 0.21875;

106x + 84y = 22.9;

y = 0.21875 – x;

106x + 84 * (0.21875 – x) = 22.9;

106x + 18.375 – 84x = 22.9;

22x = 4.525;

x = 0.2057 mol;

y = 0.01305 mol;

7. find the masses of salts:

m (Na2CO3) = 21.804 g.

m (NaHCO3) = 1.096 g.

Answer: 21.804 g Na2CO3, 1.096 g NaHCO3.