(1)CnH2n+2+ O2→nCO2+(n+1)H2O
1 1.5n+0.5 n (n+1)
据题意:1+1.5n+0.5=n+n+1 n=1,R为CH4
CnH2n-2+ O2→nCO2+(n-1)H2O
1 1.5n-0.5 n (n-1)
据题意:n+n-1-(1+1.5n-0.5)>0,n>3据题意,B为气体∴B为C4H6
(2)设炔Q为XL,消耗O2为YL
C4H6+5.5O2→4CO2+3H2O
1 5.5 4 3 Δ0.5 解得:X=0.28(L)
XL YL (2.38-2.24)L Y=1.54(L)
则烷R和O2总体积:2.24-0.28-1.54=0.42(L)
又设B中甲烷允许的最大体积为ZL
CH4~2O2
1 2 解得:(L)
ZL 2ZL
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