When burning 7.48 g of hydrogen sulfide, 4.48 liters of sulfur oxide (IV) were released.

When burning 7.48 g of hydrogen sulfide, 4.48 liters of sulfur oxide (IV) were released. Determine the mass fraction of impurities in hydrogen sulfide.

Let’s write the reaction equation:
2H2S + 3O2 = 2SO2 + 2H2O
Let us find the amount of SO2 formed by the formula n (SO2) = V (SO2) / Vm = 0.2 mol
From the coefficients of the equation we obtain that n (H2S) = n (SO2) = 0.2 mol
Let’s calculate the mass of pure hydrogen sulfide, which reacted according to the formula m (H2S) = n * M = 6.8 g
We find the mass fraction of impurities as the difference between 1 and the mass fraction of pure H2S in the mixture w (approx) = 1 – 6.8 / 7.48, or as a percentage w (approx.) = 0.09 * 100% = 9%
Answer: the mass fraction of impurities is 9%