Question

Excess dilute HCl was added to sodium sulphite 960cm3 of sulphur IV oxide has was produced calculate the mass of the sulphate used

Answer 1

Putting molar gas volume =$24000cm^3$ and molar mass of sulphite =126g, mass of the sulphite used can be calculated as follows.

number of moles of sulphur (iv) oxide $=\frac{960cm^3 \times 1mole}{24000cm^3}=0.04moles$

Equation:

$Na_2SO_3 (s)+2HCl_{aq} \to 2NaCl_{aq} +SO_2(g)+H_2O_(l)$

mole ration $Na_2SO_3:SO_2$ is 1:1

number of moles of $Na_2SO_3=0.04moles$

mass of $Na_2SO_3=126gmol^{-1} \times 0.04=5.04g$