Question

A metal ore weighing 6.125 g was roasted in air to get 5.725 g of the oxide. The oxide was then put into sulfuric acid and dried. This gave us a white solid of mass 7.725 g. The final solid is a

• A. Sulphide
• B. Sulphate
• C. Sulphite
• D. None of the above

Answer 1

The correct option is B Sulphate

Number of moles of sulphur present in the ore = $\frac{6.125-5.725}{32-16}=\frac{1}{40}$
Mass of metal = $6.125-\frac{1}{40}\times 32=5.325g$
Mass of new ion in the final white solid = $\frac{7.725-5.325}{1/40}=96g/mol$
Let's assume the formula of the ion to be $S{O}_x$
Hence, $96=32+16\times x64=16\times x⟹x=4$
Hence, the ion is $S{O}_4^{2-}$.