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
May 7, 2018

Answer:

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#.