Question

生铁和钢都是铁合金,生铁中含碳量在2.0%-4.3%,钢的含碳量在0.03%-2.0%之间,将一块质量为10.0g的铁合

Answer 1

设该合金中铁的质量为x

m(H2)=4.00L×0.0880g/L=0.352g

Fe + H2SO4 = FeSO4 + H2↑

56 2

x 0.352g

x = (56×0.352g)/2 = 9.856g

该铁合金中碳的质量分数为:[(10.00g-9.856g)/10.0g ]×100%=1.44%

介于0.03%—2.0%之间,该铁合金是钢.

设反应后所的溶液中FeSO4的质量为y

y=(152×0.352g)/2=26.75g

FeSO4的质量分数为:[6.75g/(100g+9.856g—0.352g)]×100%=24.4%