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...?

1 Answer
Mar 2, 2017

The balanced equation of the combustion reaction of #C_6H_12#

#C_6H_12+9O_2->6CO_2+6H_2O#

This equation reveals that

1 mol #C_6H_12# produces 6 mol #CO_2# on reacting with excess oxygen.

Molar mass of #C_6H_12=6xx12+12xx1=84 "g/mol"#

Molar mass of #CO2=12+2xx16=44 "g/mol"#

So theoretically 84g #C_6H_12# produces #6xx44=264 g # #CO_2#

150g #C_6H_12# produces #264/84xx150 g # #CO_2#

But the practical yield of #CO_2# being 35%

The mass #CO_2# practically produced

#=264/84xx150 g xx35%=165g#