Đáp án:
a) 4g
b) 1M
c) 850g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{CuO}} = \dfrac{{12}}{{80}} = 0,15\,mol\\
{n_{HCl}} = 0,1 \times 2 = 0,2\,mol\\
\dfrac{{0,15}}{1} > \dfrac{{0,2}}{2} \Rightarrow CuO \text{ dư}\\
{n_{CuO}} \text{ dư}= 0,15 - \frac{{0,2}}{2} = 0,05\,mol\\
{m_{CuO}}\text{ dư} = 0,05 \times 80 = 4g\\
b)\\
{n_{CuC{l_2}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{C_M}CuC{l_2} = \dfrac{{0,1}}{{0,1}} = 1M\\
c)\\
2AgN{O_3} + CuC{l_2} \to Cu{(N{O_3})_2} + 2AgCl\\
{n_{AgN{O_3}}} = 2{n_{CuC{l_2}}} = 0,2\,mol\\
{m_{{\rm{dd}}AgN{O_3}}} = \dfrac{{0,2 \times 170}}{{4\% }} = 850g
\end{array}\)
