trộn 400g dd KOH 5,6% với 300g dd CuSO4 16% khỏi lượng kết tủa thu được là bao nhiêu nêu cách tính nha mn làm giúp em vote5s ctlhn

2 câu trả lời

Em tham khảo nha :

\(\begin{array}{l}
2KOH + CuS{O_4} \to Cu{(OH)_2} + {K_2}S{O_4}\\
{n_{KOH}} = \dfrac{{400 \times 5,6\% }}{{56}} = 0,4mol\\
{n_{CuS{O_4}}} = \dfrac{{300 \times 16\% }}{{160}} = 0,3mol\\
\dfrac{{0,4}}{2} < \dfrac{{0,3}}{1} \Rightarrow CuS{O_4}\text{ dư}\\
{n_{Cu{{(OH)}_2}}} = \dfrac{{{n_{KOH}}}}{2} = 0,2mol\\
{m_{Cu{{(OH)}_2}}} = 0,2 \times 98 = 19,6g
\end{array}\)

 

Khối lượng kể tủa thu được là 19,6 gam
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