85.6g CH4 = 5.34 moles
each mole CH4 reacts with 1 mole of O2
so, you need 5.34 moles O2 = 85.39g O2
How many grams of oxygen are required to completely burn 85.6 grams of methane (CH4)?
CH4 + O2 → H2O + CO2
5 answers
oops: O2 = 32, not 16, so that's 170.78g
How many grams of 02 are needed to form 17 grams of H2O
n=M/RFM
n=number of moles RFM= relative formula mass
n=17/18=0.944
0.944 moles
then
0.944=x/16
x=0.944*16
x=15.104
15.105 grams
n=number of moles RFM= relative formula mass
n=17/18=0.944
0.944 moles
then
0.944=x/16
x=0.944*16
x=15.104
15.105 grams
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1 answer