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s344n2d4d5 [400]

2 years ago

12

Consider the reaction between nis2 and o2: 2nis2(s)+5o2(g)→2nio(s)+4so2(g) when 11.2 g of nis2 are allowed to react with 5.43 g

of o2, 4.89 g of nio are obtained.

1 answer:

stellarik [79]2 years ago

6 0

M(NiS₂) = 11.2 g.
n(NiS₂) = m(NiS₂) ÷ M(NiS₂).
n(NiS₂) = 11.2 g ÷ 122.8 g/mol.
n(NiS₂) = 0.091 mol.
m(O₂) = 5.43 g.
n(O₂) = 5.43 g ÷ 32 g/mol.
n(O₂) = 0.17 mol; limiting reactant.
From chemical reaction: n(NiS₂) : n(O₂) = 2 : 5.
0.091 mol : n(O₂) = 2 : 5.
n(O₂) = 0.2275 mol, not enough.
n(NiO) = 4.89 g .
n(O₂) : n(NiS) = 5 : 2.
n(NiS) = 0.068 mol.
m(NiS) = 0.068 mol · 74.7 g/mol = 5.08 g.
percent yield = 4.89 g / 5.08 g · 100% = 96.2%.

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