# How do you find the mass of CO2 produced in a reaction of 150.0g of C6H12 in sufficient (excess) oxygen if the reaction has a 35.00% yield?

## (3) marks are available for this question, so I'm guessing that there are about three steps to this question...?

Mar 2, 2017

The balanced equation of the combustion reaction of ${C}_{6} {H}_{12}$

${C}_{6} {H}_{12} + 9 {O}_{2} \to 6 C {O}_{2} + 6 {H}_{2} O$

This equation reveals that

1 mol ${C}_{6} {H}_{12}$ produces 6 mol $C {O}_{2}$ on reacting with excess oxygen.

Molar mass of ${C}_{6} {H}_{12} = 6 \times 12 + 12 \times 1 = 84 \text{g/mol}$

Molar mass of $C O 2 = 12 + 2 \times 16 = 44 \text{g/mol}$

So theoretically 84g ${C}_{6} {H}_{12}$ produces $6 \times 44 = 264 g$ $C {O}_{2}$

150g ${C}_{6} {H}_{12}$ produces $\frac{264}{84} \times 150 g$ $C {O}_{2}$

But the practical yield of $C {O}_{2}$ being 35%

The mass $C {O}_{2}$ practically produced

=264/84xx150 g xx35%=165g