Question

Calculate the mass of CO2 that can be produced if the reaction of 46.8 g of pentane and sufficient oxygen has a 58.0 %yield. C5H12(g)+8O2(g)→5CO2(g)+6H2O(g)

Answer 1

Step 1

The reaction must be written and balanced:

C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(g)

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Step 2

Information provided:

The limiting reactant is pentane.

% yield = Actual yield/theoretical yield x 100 = 58.0 %

Data needed:

The molar mass of pentane (C5H12) = 72.1 g/mol

The molar mass of CO2 = 44.0 g/mol

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Step 3

By stoichiometry the mass of CO2 is determined:

1 mol pentane = 72.1 g

1 mol CO2 = 44.0 g

72.1 g C5H12 ------- 5 x 44.0 g CO2

46.8 g C5H1------- X

X = 246.8 g C5H1 x 25 x 44.0 g CO/72.1 g 2C5H1

X = 142.8 g CO2 = theoretical yield

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Step 4

% yield = actual yield/theoretical yield x 100 = 58.0 %

Actual yield = 58%/100 x theoretical yield = 58%/100 x 142.8 g = 82.8 g CO2

Answer: 82.8 g CO2 produced2