Question #17a18

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Answer 1 · 2018-01-29T02:04:14

The mass of $CO_2$ produced will be 62.7 g

First, we use the ideal gas law to determine the number of moles of methane:

$PV=nRT$

which we will write as

$n=(PV)/(RT) = (749 xx 35.0)/(62.36 xx295) = 1.425 "mol"$

Now, the balanced equation for the reaction:

$CH_4 + 2 O_2 rarr CO_2 + 2H_2O$

Since the ration of $CH_4$ to $CO_2$ is one-to-one, there will be 1.425 mol of $CO_2$ produced, with a mass of

$1.425 "mol" xx 44g/"mol"=62.7 g$