A 60 g sample of pyrite ore is 80% FeS2 by mass (Fe = 56, S = 32, H = 1, O = 16). Roasting gives SO2 (all S converted), which is converted into H2SO4 via SO2 -> SO3 -> H2SO4 with an overall 90% yield from SO2. Find the mass of H2SO4 in grams, to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3mc051

A $60 g$ sample of pyrite ore is $80%$ $FeS X_{2}$ by mass (Fe = 56, S = 32, H = 1, O = 16). Roasting gives $SO X_{2}$ (all S converted), which is converted into $H X_{2} SO X_{4}$ via $SO X_{2} SO X_{3} H X_{2} SO X_{4}$ with an overall $90%$ yield from $SO X_{2}$ . Find the mass of $H X_{2} SO X_{4}$ in grams, to two decimal places.

Numerical answer type, answer in g, tolerance 0.005.

FeS X_{2}

= 0.80 \times 60 = 48

= 48/120 = 0.40

mol. Each

FeS X_{2}

gives

2

S, so

2 \times 0.40 = 0.80

mol

SO X_{2}

. At

90%

yield,

H X_{2} SO X_{4}

= 0.80 \times 0.90 = 0.72

mol, mass

= 0.72 \times 98 = 70.56 g