How do you determine the mass of carbon dioxide produced when 0.85 g of butane reacts with oxygen according to the following equation: 2C4H10+13O2→8CO210H2O?

1 Answer
anor277
Sep 6, 2016

C4H10(g)+132O2(g)→4CO2(g)+5H2O(l)

5.58×10−2moles of CO2 are produced.

Explanation:

Moles of butane = 0.85⋅g58.12⋅g⋅mol−1 = 1.46×10−2mol.

Given the stoichiometry, each moles of butane gives 4⋅mol CO2 upon complete combustion.

And thus moles of CO2 = 4×1.46×10−2mol=5.58×10−2mol.

This is an equivalent mass of 5.58×10−2mol×44.01⋅g⋅mol−1≅2.5⋅g.

In my stoichiometric equation, I used an half equiv of dioxygen in order to balance the equation. I do this because it makes the 'rithmetic a bit easier.

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