Hòa tan hoàn toàn 1,44g kim loại hóa trị II bằng 250ml dung dịch H2SO4 0,3M. Để trung hòa lượng axit dư cần dùng 60ml dung dịch NaOH 0,5M. Đó là kim loại gì?

1 câu trả lời

Gọi kim loại có hoá trị II là M

Đổi 250ml= 0,25 l

       60ml= 0,06 l

Ta có PT:  M + H2SO4 -> MSO4 + H2 (1)
H2SO4 + 2NaOH -> NA2SO4 + 2H2O (2)
nH2SO4 = CM .V= 0,3*0,25= 0.075 mol
nNaOH = Cm. V= 0,5*0,06= 0.03 mol
Từ (2) => nNaOH=2nH2SO4 

nH2SO4 dư = 0.015 mol => nH2SO4(1) = 0.075 - 0.015 = 0.06 mol
Từ (1) => nM = 0.06 => M = m/n= 1.44/0.06 = 24 =>M là Mg

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