# Question #e21bc

Nov 23, 2015

5.90 g $C {H}_{4}$

#### Explanation:

To solve this kind of problem you just set up a dimensional analysis table. Since we are given the mass of the $C {O}_{2}$ we should start there.

$64.8 g C {O}_{2}$ x $\frac{1 m o l C {O}_{2}}{44 g C {O}_{2}}$ x $\frac{2 m o l C {H}_{4}}{8 m o l C {O}_{2}}$ x $\frac{16.03 g C {H}_{4}}{1 m o l C {H}_{4}}$ = $5.90 g C {H}_{4}$

One important thing to remember when solving these kind of stoichiometry problems is paying attention to the mole ratio when going from one species to the other.
In this problem the mole ratio is between $C {H}_{4}$ and $C {O}_{2}$. You can see that the ratio is the third portion of the analysis.

I hope this helps!