Pyrite (FeS2) of 80.0% purity is used to make sulphuric acid. Each mole of sulphur in the ore is converted to one mole of H2SO4 with an overall yield of 90.0%. Mass of H2SO4 (in g, two decimals) obtainable from 60.0 g of the ore is (Fe = 56, S = 32, H = 1, O = 16).

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JEE Advanced levelNumerical answerDifficulty 3 of 3pn165

Pyrite ( $FeS X_{2}$ ) of $80.0%$ purity is used to make sulphuric acid. Each mole of sulphur in the ore is converted to one mole of $H X_{2} SO X_{4}$ with an overall yield of $90.0%$ . Mass of $H X_{2} SO X_{4}$ (in g, two decimals) obtainable from $60.0 g$ of the ore is (Fe = 56, S = 32, H = 1, O = 16).

Numerical answer type, answer in g, tolerance 0.01.

Answer: 70.56 g

Pure FeS2 = 60.0 x 0.800 = 48.0 g; molar mass 120, so n = 0.40 mol. Each FeS2 supplies 2 S, so 2 x 0.40 = 0.80 mol S. With 90.0% yield n(H2SO4) = 0.720 mol. Mass = 0.720 x 98 = 70.56 g.

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Pyrite (FeSX2​) of 80.0% purity is used to make sulphuric acid. Each mole of sulphur in the ore is converted to one mole of HX2​SOX4​ with an overall yield of 90.0%. Mass of HX2​SOX4​ (in g, two decimals) obtainable from 60.0 g of the ore is (Fe = 56, S = 32, H = 1, O = 16).

Answer: 70.56 g

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