Most combustion reactions occur in excess of `O_(2)` (i.e., more than enough `O_(2)` to burn the substance completely). Calculate the mass of `CO_(2)` (in grams) produced by buring `4.00 mol` of `CH_(4)` in excess `O_(2)`.

Strategy: using the balanced equation
`{:(CH_(4), + 2O_(2), rarr CO_(2), + 2H_(2) O),(1mol, 2mol,1 mol,2mol),(16.0 g,2(32.0)g,44.0g,2(18.0g)):}`
we find out that 1 mol of `CH_(4)` is chemically equivalent to 1 mol of `CO_(2)` or `44.0g` of `CO_(2)`.

Explanation

According to the equation, `n_(CO_(2))`, produced is equal to `n_(CH)` used. Thus, 4 mol `CH_(4)` produces 4 mol `CO_(2)`. The molar mass of `CO_(2)` is `44g`. Thus, the mass of 4 mol `CO_(2)` is `4xx44 = 176g`.
Alternatively, using the right unit factors,
`?g CO_(2) = 4.00 "mol" CH_(4) xx (1 "mol" CO_(2))/(1 "mol" CH_(4)) xx (44.0 g CO_(2))/(1 "mol" CO_(2))`
`= 1.76xx10^(2) g CO_(2)`