Question

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...?

Answer 1

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`