Question

Question #e21bc

Answer 1

5.90 g `CH_4`

Explanation:

To solve this kind of problem you just set up a dimensional analysis table. Since we are given the mass of the `CO_2` we should start there.

`64.8 g CO_2` x `(1 molCO_2)/(44 g CO_2)` x `(2 mol CH_4)/(8mol CO_2)` x `(16.03g CH_4)/(1 mol CH_4)` = `5.90 g CH_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 `CH_4` and `CO_2`. You can see that the ratio is the third portion of the analysis.

I hope this helps!