Em tham khảo nha :
\(\begin{array}{l}
1)\\
2NaOH + CuS{O_4} \to Cu{(OH)_2} + N{a_2}S{O_4}\\
{n_{NaOH}} = {C_M} \times V = 0,3 \times 0,5 = 0,15mol\\
{n_{CuS{O_4}}} = \dfrac{m}{M} = \dfrac{{16}}{{160}} = 0,1mol\\
\dfrac{{0,15}}{2} < \dfrac{{0,1}}{1} \Rightarrow CuS{O_4}\text{ dư}\\
{n_{Cu{{(OH)}_2}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,075mol\\
{m_{Cu{{(OH)}_2}}} = n \times M = 0,075 \times 98 = 7,35g\\
2)\\
AgN{O_3} + HCl \to AgCl + HN{O_3}\\
{n_{AgN{O_3}}} = {C_M} \times V = 0,15 \times 1 = 0,15mol\\
{n_{AgCl}} = {n_{AgN{O_3}}} = 0,15mol\\
{m_{AgCl}} = n \times M = 0,15 \times 143,5 = 21,525g\\
{n_{HCl}} = {n_{AgN{O_3}}} = 0,15mol\\
{C_{{M_{HCl}}}} = \dfrac{n}{V} = \dfrac{{0,15}}{{0,1}} = 1,5M
\end{array}\)
