# 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?

##### 1 Answer

#### Answer:

Given that we are told only the mass of

Mass of

Mass of

#### Explanation:

First step is always to create a balanced chemical reaction:

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

The number of moles of

Each mole of

By the same reasoning for