1)СH4+2O2=CO2+2H2O
V(CH4)=5/16=0,3125моль
n(O2)=0,625 моль
V(O2)=0,625*22,4=14 л
2)2С2H6+7O2=4CO2+6H2O
n(C2H6)=2/30=0,067моль
n(O2)=0,2345 моль
V(O2)=0,2345*22,4=5,2528 л
V(O2)общ=14+5,2528 =19,2528 л
3) 19,2528 л-100%
x-21%
x=4 л
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
1)СH4+2O2=CO2+2H2O
V(CH4)=5/16=0,3125моль
n(O2)=0,625 моль
V(O2)=0,625*22,4=14 л
2)2С2H6+7O2=4CO2+6H2O
n(C2H6)=2/30=0,067моль
n(O2)=0,2345 моль
V(O2)=0,2345*22,4=5,2528 л
V(O2)общ=14+5,2528 =19,2528 л
3) 19,2528 л-100%
x-21%
x=4 л