Question

The synthesis of maleic acid anhydride (C₄H₂O₃) can be accomplished by reacting benzene (C₆H₆) and oxygen gas in the following chemical reaction:

2 C₆H₆(l) + 9 O₂(g) → 2 C₄H₂O₃(s) + 4 CO₂(g) + 4 H₂O(g)

What is the mass in grams of benzene that is required to produce 109.5 grams of maleic acid anhydride assuming the reaction has a 72.6% yield?

Answer 1

120.3 g of benzene is required to produce 109.5 grams of maleic acid anhydride.

The reaction equation is;

2 C₆H₆(l) + 9 O₂(g) → 2 C₄H₂O₃(s) + 4 CO₂(g) + 4 H₂O(g)

Actual yield of   maleic anhydride = 109.5 g

Theoretical yield = 100 × Actual yield/% yield

Theoretical yield =  100 ×  109.5/72.6

Theoretical yield  = 150.8 g

Number of moles of maleic anhydride = mass/molar mass = 150.8 g/98.06 g/mol = 1.54 moles

From the reaction equation;

2 moles of benzene yields 2 moles of maleic anhydride

1.54 moles of benzene yields 1.54 moles of maleic anhydride

Mass of benzene required = 1.54 moles × 78.11 g/mol

Mass of benzene required = 120.3 g of benzene.

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