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Anonymous asked in Science & MathematicsChemistry · 1 month ago

Consider the unbalanced reaction NiS2(s) + O2(g) --> NiO(s) +SO2(g) When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained.?

Consider the unbalanced reaction NiS2(s) + O2(g) --> NiO(s) + SO2(g)

When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. The theoretical yield of NiO is_ the limiting reactant is_and the percent yield is_

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  • 1 month ago

    2 NiS2 + 5 O2 → 2 NiO + 4 SO2

    (11.2 g NiS2) / (122.8234 g NiS2/mol) = 0.0911878 mol NiS2

    (5.43 g O2) / (31.99886 g O2/mol) = 0.169694 mol O2

    0.169694 moles of O2 would react completely with 0.169694 x (2/5) = 0.0678776 mole of NiS2, but there is more NiS2 present than that, so NiS2 is in excess and

    O2 is the limiting reactant.

    (0.169694 mol O2) x (2 mol NiO / 5 mol O2) x (74.6928 g NiO/mol) = 5.069968 g =

    5.07g NiO in theory

    (4.86 g) / (5.069968 g) = 0.95859 = 95.9% yield NiO

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