The equation for the complete combustion of methane is:
CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l)
To calculate the number of grams of CO2 produced by the reaction of
50.6 g of methane, the first conversion factor to use is:
a. 1 mol CH4/16.0 g CH4
b. 2 mol O2/1 mol CO2
c. 16.0 g CH4/1 mol CO2
d. 44.0 g CO2/2 mol CO2

Hi, teatea-ga:

The "first" conversion factor to use is A. 1 mole of methane per 16.0
grams of methane (adding the atomic weight 12 for carbon plus four
times atomic weight 1 for hydrogen, roughly).

Multiplying 50.6 grams (of methane) by this conversion factor tells us
how many moles of methane are available for the combustion reaction. 
Since the reaction produces an equal number of moles of carbon dioxide
as the number of moles of methane present (as can be seen easily from
the equal number of carbon atoms accounted for), once we do that first
conversion it will be a simple matter to perform a second conversion
(getting the number of grams of CO2 from the number of moles).  For
example:

50.6 g * (1/16.0) mol/g = 3.16 moles of methane

Then secondarily the equal 3.16 moles of carbon dioxide would get
converted to grams by multiplying by the molecular weight of CO2 (but
the question doesn't ask us to carry this out).

The answer is A. 1 mol CH4/16.0 g CH4.

regards, mathtalk-ga