1.3A + B =xC + 2D
3 1 2
c(A耗) c(B耗) 0.5mol/L
c(A耗)=0.75mol/L c(B耗)=0.25mol/L
则c(A余)+0.75mol/L =5/3c(A余)+0.25mol/L
c(A余)=0.75mol/L
2.v(D)=0.5/5=0.1mol/(Lmin)
v(B)=1/2 v(D)=0.05mol/(Lmin)
3.v(C)=0.1mol/(L*min)
v(D):v(C)= 1 则 x=2
1.3A + B =xC + 2D
3 1 2
c(A耗) c(B耗) 0.5mol/L
c(A耗)=0.75mol/L c(B耗)=0.25mol/L
则c(A余)+0.75mol/L =5/3c(A余)+0.25mol/L
c(A余)=0.75mol/L
2.v(D)=0.5/5=0.1mol/(Lmin)
v(B)=1/2 v(D)=0.05mol/(Lmin)
3.v(C)=0.1mol/(L*min)
v(D):v(C)= 1 则 x=2