Cu + 4HNO3(浓) == Cu(NO3)2 + 2NO2↑ + 2H2O
3Cu + 8HNO3(稀) == 3Cu(NO3)2 + 2NO↑ + 4H2O
收集到的气体为0.05mol,设NO2为x,NO为y,则:
x+y=0.05
x/2+3y/2=0.03
x=0.045 y=0.005
所以消耗的HNO3的物质的量是:
0.045*2+0.005*2=0.11mol
3NO2+H2O=2HNO3+NO
4NO+3O2+2H2O=4HNO3
消耗的O2量为:
(0.045/3+0.005)*3*22.4/4=0.336L