解析:1.92g铜粉,n=m/M=1.92/64=0.03mol
NO2和NO组成的混合气体1.12L n=V/Vm=1.12/22.4=0.05mol
设NO和NO2物质的量分别为n1和n2 则有:
e-守恒:0.03*2 = n1*3 + n2*1 ------------1
n1+ n2= 0.05---------------------2
由1,2式求得:n1=0.005mol n2=0.045mol
V(NO)= n1*Vm= 0.005* 22.4=0.112(L) 答:略
解析:1.92g铜粉,n=m/M=1.92/64=0.03mol
NO2和NO组成的混合气体1.12L n=V/Vm=1.12/22.4=0.05mol
设NO和NO2物质的量分别为n1和n2 则有:
e-守恒:0.03*2 = n1*3 + n2*1 ------------1
n1+ n2= 0.05---------------------2
由1,2式求得:n1=0.005mol n2=0.045mol
V(NO)= n1*Vm= 0.005* 22.4=0.112(L) 答:略