10mL混合酸中含有:n(H +)=0.01L×2×4mol/L+0.01L×2mol/L=0.1mol,n(NO 3 -)=0.01L×2mol/L=0.02mol,
由于铁过量,则发生反应:3Fe+2NO 3 -+8H +=3Fe 2++2NO↑+4H 2O,Fe+2H +=Fe 2++H 2↑,则
3Fe+2NO 3 -+8H +=3Fe 2++2NO↑+4H 2O
0.02mol 0.08mol 0.02mol
反应后剩余n(H +)=0.1mol-0.08mol=0.02mol,
Fe+2H +=Fe 2++H 2↑
0.02mol 0.01mol
所以:n(NO)+n(H 2)=0.02mol+0.01mol=0.03mol,
V(NO)+V(H 2)=0.03mol×22.4L/mol=0.672L,
故选B.