Find the volume of sulfur dioxide released during the interaction of a 10% sodium

Find the volume of sulfur dioxide released during the interaction of a 10% sodium sulfite solution weighing 200 g with hydrochloric acid.

1. Let’s compose the equation of the proceeding reaction:

Na2SO3 + 2HCl = 2NaCl + SO2 ↑ + H2O;

2. find the mass of sodium sulfite:

m (Na2SO3) = w (Na2SO3) * m (solution) = 0.1 * 200 = 20 g;

3.Calculate the chemical amount of sulfite:

n (Na2SO3) = m (Na2SO3): M (Na2SO3);

M (Na2SO3) = 2 * 23 + 32 + 3 * 16 = 126 g / mol;

n (Na2SO3) = 20: 126 = 0.1587 mol;

4.determine the amount of gas released:

n (SO2) = n (Na2SO3) = 0.1587 mol;

5.calculate the volume of sulfur dioxide:

V (SO2) = n (SO2) * Vm = 0.1587 * 22.4 = 3.55 liters.

Answer: 3.55 liters.