# After 240 g of a 25% sodium hydroxide solution, 33.6 liters of carbon dioxide were passed. What is the mass fraction of NaHCO3

Decision:

M (NaOH) = 40 g / mol;

VM = 22.4 liters

m (NaOH) = [m (solution). W%] / 100% = (240. 25) / 100 = 60 g.

Then

m (H2O) = m (p-pa) – m (NaOH) = 240 – 60 = 180 g.

n (NaOH) = m (NaOH) / M (NaOH) = 60/40 = 1.5 mol.

n (CO2) = V / VM = 33.6 / 22.4 = 1.5 mol.

The reaction equation has the form:

NaOH + CO2 ↔ NaHCO3

It follows from the equation that 1 mol of CO2 reacts with 1 mol of NaOH and in this case 1 mol of NaHCO3 is formed:

n (NaOH) = n (NaOH) = n (NaHCO3) = 1.5 mol.

So, in this case, all carbon dioxide and sodium hydroxide completely reacted, and only sodium bicarbonate is present in the solution.

Calculate the mass of NaHCO3, we get:

m (NaHCO3) = M (NaHCO3). n (NaHCO3) = 84. 1.5 = 126 g.

We find the mass of the resulting solution, which will be the remaining water and sodium bicarbonate, we obtain:

m (solution) = m (H2O) + m (NaHCO3) = 180 + 126 + 306 g.

w% (NaHCO3) = (126.100) / 306 = 41.18%.

Answer: w% (NaHCO3) = 41.18%.