Answer to Question #234674 in Inorganic Chemistry for Lee

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