Question

with the following balanced equation 2C4H10 + 13O2 = 8CO2 + 10H2OHow many grams of CO2 are produced from the combustion of 100 grams of butane?

Answer 1

Step 1

The reaction provided:

2 C4H10 + 13 O2 = 8 CO2 + 10 H2O (completed and balanced)

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Step 2

Information provided:

100 g of butane (C4H10)

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Information needed:

The molar masses of:

C4H10) 58.12 g/mol

CO2) 44.01 g/mol

(the periodic table is helpful here)

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Step 3

By stoichiometry,

2 C4H10 + 13 O2 = 8 CO2 + 10 H2O

2 x 58.12 g C4H10 ------------ 8 x 44.01 g CO2

100 g C4H10 ------------ X

X = 100 g C4H10 x 8 x 44.01 g CO2/2 x 58.12 g C4H10

X = 303 g approx.

Answer: 303 g of CO2 produced