Question

The hydrocarbon, pentane, C5H12, reacts with oxygen to form carbon dioxide and water. In a 3.00...

The hydrocarbon, pentane, C5H12, reacts with oxygen to form carbon dioxide and water. In a 3.00 L container at 25 ºC are placed 0.520 moles of O2 gas and 0.400 moles of solid C5H12.

b) How many moles of CO2(g) is produced?

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Answer #1

Balanced chemical equation is:

C5H12 + 8 O2 ---> 5 CO2 + 6 H2O

1 mol of C5H12 reacts with 8 mol of O2

for 0.4 mol of C5H12, 3.2 mol of O2 is required

But we have 0.52 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

According to balanced equation

mol of CO2 formed = (5/8)* moles of O2

= (5/8)*0.52

= 0.325 mol

Answer: 0.325 mol

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