Find the theoretical yield of silicon monocarbide if 50.0 grams of silicon dioxide reacts with carbon. then, calculate the actual amount of silicon monocarbide produced in the lab if the percent yield was 87.2%. (work backwards to find actual yield.)

1) Answer is: yield of silicon monocarbide is 33,41 g.
Chemical reaction: SiO₂  + 3C  → SiC  + 2CO.
m(SiO₂) = 50,0 g.
n(SiO₂) = m(SiO₂) ÷ M(SiO₂).
n(SiO₂) = 50 g ÷ 60 g/mol.
n(SiO₂) = 0,833 mol; limiting reagent.
Missing question: m(C) = 79,1 g.
n(C) = 79,1 g ÷ 12 g/mol = 6,59 mol.
From chemical reaction: n(SiO₂) : n(SiC) = 1 : 1.
n(SiC) = 0,833 mol.
m(SiC) = 0,833 mol · 40,11 g/mol = 33,41 g.

2) Answer is: actual yield is 29,13 g.
Chemical reaction: SiO₂  + 3C  → SiC  + 2CO.
Percent yield = 87,2% ÷ 100% = 0,872.
Percent yield = actual yield / theoretical yield.
m(SiC) = 33,41 g ·0,872.
m(SiC) = 29,13 g.
Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.

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