# What is the maximum mass of S8 that can be produced by combining 91.0g of each reactant?

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What is the maximum mass of S8 that can be produced by combining 91.0g of each reactant?

8SO2+16H2S------->3S8+16H2O

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8 SO2 + 16 H2S → 3 S8 + 16 H2O

(91.0 g SO2) / (64.0644 g SO2/mol) = 1.4204 mol SO2

(91.0 g H2S) / (34.0814 g H2S/mol) = 2.6701 mol H2S

2.67 mole of H2S would react completely with 2.67 x (3/16) = 0.50 mol SO2, but there is more SO2 present than that, so SO2 is in excess and H2S is the limiting reactant.

(2.6701 mol H2S) x (3/16) x (256.5240 g S8/mol) = 128 g S8

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Molar mass of SO2 is sixty 4.a million g/mol so 89.0 g is 89.0/sixty 4.a million mol = a million.39 mol Molar mass of H2S is 34.a million g/mol so 89.0 g is 89.0/34.a million mol = 2.sixty one mol in accordance to the equation, 8 moles of SO2 require sixteen moles of H2S, i.e. a million mole of SO2 calls for 2 moles of H2S. as a result, a million.39 moles of SO2 calls for 2.seventy 8 moles of H2S. yet purely 2.sixty one moles of H2S are contemporary so the H2S is the proscribing reagent. in accordance to the equation, sixteen moles of H2S produce 3 moles of S8. So a million mole of H2S will produce 3/sixteen moles of S8 and a couple of.sixty one moles of H2S will produce 2.sixty one x 3/sixteen moles of S8 = 0.40 9 moles of S8. Molar mass of S8 is 256.5 g/mol so 0.40 six moles of S8 is 0.40 9 x 256.5 g = one hundred twenty five.7 g S8

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