Question

When octane (C8H18; MW = 114.0 g/mol)) is burned in a particular internal combustion engine, the yield of carbon dioxide is 93%. What mass of carbon dioxide will be produced in this engine when 15.0 g of octane is burned with 15.0 g of oxygen gas?

Answer 1

First you need to write and balance a reaction for the process.
Then you need to find the limiting reactant. This is done by dividing the amount of moles by it's respective stoichiometric coefficient (from the balanced reaction) and comparing to see which has less. The one with less is the limiting reactant.

Then proceed with the moles of the limiting reactant (before dividing) to find the moles of product. Set up a ratio using the species of interest, like so:

e.g. for a general reaction:

\(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients ,

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

From here you can isolate what you need.
For example: if you have 2 moles of B, how many moles of C can you produce?

solve algebraically: \(\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)
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To convert mass to moles, use the relationship:

\(n=\dfrac{m}{M}\)

where, M=molar mass, m=mass, and n= moles.

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