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2C2H6+7O2->4CO2+6H2O
pV=nRT|:RT
n=pV/RT
n=1000*10/8,314*273,15
n=10000/2271
n=4,4 mola
Z równania reakcji wiemy, że:
2 mole C2H6 - 7 moli O2
4,4 mole - X
X=15,4 mol
pV=nRT|:p
V=nRT/p
V=15,4*8,314*273,15
V=35 dm3 tlenu
21% - 35dm3
100% - X
X=166,54 dm^3 powietrza