CnH2n+2
w(C)=12n/(12n+2n+2)*100%
0.8276/1=12n/(14n+2)
n=4
C4H10
Mg+S>(=)MgS
Mg+H2O>MgO+H2
Mg+CuCl2>MgCl2+Cu
Mg+Cl2>MgCl2
Mg+O2>2MgO
Mg+H2SO4=MgSO4+H2O
M(C)=62,5г
w(C)=96\%
V(CO2)-?
Решение:
С+О2-->СО2
m(C)=62,5*0,96=60г
n(C)=m(C)/M(C)=60/12=5моль
V(CO2)=n(C)*Vm=5*22,4=112л
LiOH
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