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mZn(OH)₂=99,4u
ZnSO₄ + 2NaOH ---> Zn(OH)₂ + Na₂SO₄
161,4g soli ----- 2mole NaOH
16,14g soli ------ xmoli NaOH
x = 0,2mola NaOH
Nadmiar NaOH = 0,25-0,2 = 0,05mola
161,4g soli ----- 99,4g osadu
16,14g soli ------ xg osadu
x = 9,94g osadu
Zn(OH)₂ + 2NaOH ---> Na₂[Zn(OH)₄]
99,4g osadu ----- 2mole NaOH
xg osadu --------- 0,05mola NaOH
x = 2,485g osadu
Pozostało: 9,94g-2,485g = 7,455g osadu