Asked by Anonymous
If 45L of methane undergoes complete combustion at 1.1atm and 95 ° F how many grams of each product are formed?
Answers
Answered by
bobpursley
CH4 + 2O2 >> CO2 + 2H2O
moles methane:
n=PV/RT go through that, convert temps to Kelvins first. Use the correct R.
then, moles CO2=n, moles water vapor=2n
convert each to grams: mass=moles*molmasseach
moles methane:
n=PV/RT go through that, convert temps to Kelvins first. Use the correct R.
then, moles CO2=n, moles water vapor=2n
convert each to grams: mass=moles*molmasseach
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