Question

把表面被氧化的镁条4g放入100g稀硫酸中,恰好完全反应,生成0.2g气体

Answer 1

(1)

Mg+H2SO4==MgSO4+H2↑

24 98 120 2

X Y Z 0.2g

24/X = 98/Y = 120/Z = 2/0.2g

X = 2.4g,Y = 9.8g,Z = 12g

镁的质量分数 = 2.4/4 = 60%

(2)

m(MgO) = 4-2.4 = 1.6g

MgO+H2SO4==MgSO4+H2O

40 98 120

1.6g a b

40/1.6g = 98所以/a = 120/b

a = 3.92g,b = 4.8g

所以稀硫酸的质量分数 = (Y+a)/100 = (9.8+3.92)/100 = 13.72%

(3)

m(MgSO4) = Z+b = 16.8g

反应后溶液质量 = 4+100-0.2 = 103.8g

所以反应后溶液的质量分数 = 16.8/103.8 = 16.2%