# 5 grams of Methane (CH3OH) react with 10 grams of oxygen (O2), how many grams of carbon dioxide (CO2)will be formed?

NOTE: ##CH_3OH## is methanol and ##CH_4## is methane.

I will solve for both since both undergo combustion reaction.

So,
1) ##5## ##g## of ##CH_3OH##. The molecular mass of ##CH_3OH## is
##12+3+16+1=32## ##g## ##mol^-1##
So, the moles of ##CH_3OH## are ##5/32=0.15625 ## ##mol##
##10## ##g## of ##O_2## has ##10/32=0.3125## ##mol##

The combustion of methanol produces ##CO_2## and ##H_2O##

So, ##2CH_3OH+3O_2->2CO_2+4H_2O##

Hence, ##2## ##mol## of methanol requires ##3## ##mol## oxygen to undergo complete reaction.
So, ##0.15625## ##mol## of methanol will require ##0.234375## ##mol## oxygen to undergo complete reaction.

Since, amount of oxygen supplied is in excess of what is required entire amt. of methanol will be exhausted.

From the equation,
##2## ##mol## of methanol produces ##2## ##mol## carbon dioxide
##0.15625## ##mol## of methanol will produce ##0.15625## ##mol## of carbon dioxide.

The mass of carbon dioxide##=44xx0.15625=6.875## ##g##

2) ##5## ##g## of ##CH_4##. The molecular mass of ##CH_3OH## is
##12+4=16## ##g## ##mol^-1##
So, the moles of ##CH_4## are ##5/16=0.3125 ## ##mol##
##10## ##g## of ##O_2## has ##10/32=0.3125## ##mol##

The combustion of methane produces ##CO_2## and ##H_2O##

So, ##CH_4+2O_2->CO_2+2H_2O##

Hence, ##1## ##mol## of methane requires ##2## ##mol## oxygen to undergo complete reaction.
So, ##0.3125## ##mol## of methanol will require ##0.625## ##mol## oxygen to undergo complete reaction.

Since, amount of oxygen supplied is less than what is required entire amt. of methanol will not be used up. So, here oxygen is the limiting reactant. The amt. of oxygen available will dictate the amt. of carbon dioxide produced.

From the equation,
##2## ##mol## of oxygen produces ##1## ##mol## carbon dioxide
##0.3125## ##mol## of oxygen will produce ##0.15625## ##mol## of carbon dioxide.

The mass of carbon dioxide##=44xx0.15625=6.875## ##g##

I hope that was helpful. Please take care while naming chemical .

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