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#CH_4+2O_2rarrCO_2+2H_2O#
Imole #rarr#2 moles
Convert to grams.
( #A_rC=12,A_rH=1,A_rO=16)#
#[12+(4xx1)]grarr2xx[16+(2xx1)]g#
#16grarr2xx18=36g#
I have assumed the mass of methane to be #5xx10^3g#.
So #1grarr36/16g#
So #5xx10^(3)grarr(36)/(16)xx5xx10^(3)g#
#=11.25xx10^(3)g#
#=11.25kg#