Question

COMPLETE COMBUSTION of `C_(14)H_(16)O_3` to `CO_2` and `H_2O`. You have `315.0000` grams of `C_(14)H_(16)O_3`. What is the limiting reagent? How many grams of `CO_2` are generated? How many grams of `H_2O` are generated?

Answer 1

Given that we are told only the mass of `C_(14)H_(16)O_3`, we can assume that it undergoes combustion in excess oxygen, so `C_(14)H_(16)O_3` is the limiting reagent.

Mass of `CO_2 = 836.5` `g`
Mass of `H_2O = 195.6` `g`

Explanation:

First step is always to create a balanced chemical reaction:

`C_(14)H_(16)O_3 + 33/2` `O_2 -> 14` `CO_2 + 8` `H_2O`

If you're uncomfortable with a fractional coefficient the whole equation could be multiplied by 2, but it's not necessary.

Now, the molar mass of `C_(14)H_(16)O_3` is `232` `gmol^-1`.

The number of moles of `C_(14)H_(16)O_3` is `n=m/M=315/232=1.358` `mol`

Each mole of `C_(14)H_(16)O_3` produces 14 mole of `CO_2`, and `CO_2` has a molar mass of `44` `gmol^-1`, so the mass of `CO_2` produced is:

`1.358xx14xx44=836.5` `g` of `CO_2`.

By the same reasoning for `H_2O` with its molar mass of `18` `gmol^-1`:

`1.358xx8xx18=195.6` `g` of `H_2O`.