Trộn 400 gam dung dịch KOH 5,6% với 300 gam dung dịch CuSO₄ 16%. Khối lượng kết tủa thu được là:

\({{m}_{KOH}}=\frac{{{m}_{dd\text{ }KOH}}.\text{ }C%}{100%}=\frac{400.\text{ }5,6%}{100%}=22,4\,(gam)\)

=> \({{n}_{KOH}}=\frac{{{m}_{KOH}}}{{{M}_{KOH}}}=\frac{22,4}{39+16+1}=0,4\text{ }mol\)

\({{m}_{CuS{{O}_{4}}}}=\frac{{{m}_{dd\text{ }CuS{{O}_{4}}}}.\text{ }C%}{100%}=\frac{300.16%}{100%}=48\,(gam)\)

=> \({{n}_{CuS{{O}_{4}}}}=\frac{{{m}_{CuS{{O}_{4}}}}}{{{M}_{CuS{{O}_{4}}}}}=\frac{48}{64+32+64}=0,3\text{ }mol\)

PTHH:        2KOH + CuSO₄ → Cu(OH)₂ + K₂SO₄

                        2               1               1

                       0,4            0,3            ? mol

Từ phương trình ta có tỉ lệ \(\frac{{{n}_{CuS{{O}_{4}}}}}{1}>\frac{{{n}_{KOH}}}{2}(0,3>0,2)\)

=> \({{n}_{Cu{{\left( OH \right)}_{2}}}}\text{=}\frac{{{n}_{KOH}}}{2}=0,2\text{ }mol\)

=> \({{m}_{Cu{{\left( OH \right)}_{2}}}}={{n}_{Cu{{\left( OH \right)}_{2}}}}.\text{ }{{M}_{Cu{{\left( OH \right)}_{2}}}}=0,2.\left( 64+2+32 \right)=19,6\,(gam)\)