Question

向15gFe和Fe2O3混合物中,加入150mL稀硫酸,在标况下放出1.68LH2,

Answer 1

Fe + H2SO4 = FeSO4 + H2

n(H2)=1.68/22.4 = 0.075 mol

m(Fe)=56*0.075=4.2 g

Fe2O3 + 3H2SO4 = Fe2(SO4)3 + 3H2O (设:Fe2O3为xmol)

Fe + 2Fe3+ = 3Fe2+

n(Fe3+)=2x 则n(Fe)=x

56x + 160x = 15-4.2 = 10.8 故x= 0.05 mol

m(Fe2O3)=0.05*160=8(g) ,m(Fe)=15-8=7(g)

参加反应的n(H2SO4)=0.075+3x=0.225mol

剩余的n(H2SO4)=3*0.2/2=0.3mol

总n(H2SO4)=0.225+0.3=0.525mol

c(H2SO4)=0.525/0.15=3.5(mol/L)