`1,`

`n_(AgNO_3)=\frac{50.17%}{170}=0,05(mol)`

`AgNO_3+HCl->AgCl+HNO_3`

`0,05`            `0,05`    `0,05`    `0,05`

`a,`

`m_(AgCl)=0,05.143,5=7,175(g)`

`b,`

`C%_(HCl)=\frac{0,05.36,5}{20}.100=9,125%`

`c,`

`m_(dd)=50+20-7,175=62,825(g)`

`C%_(HNO_3)=\frac{0,05.63}{62,825}.100=5,01%`

`2,`

`n_(FeCl_3)=\frac{50.16,25%}{162,5}=0,05(mol)`

 `FeCl_3+3KOH->Fe(OH)_3+3KCl`

`0,05`          `0,15`         `0,05`        `0,15`

`a,`

`m_(Fe(OH)_3)=0,05.107=5,35(g)`

`b,`

`C%_(KOH)=\frac{0,15.56}{50}.100=16,8%`

`c,`

`m_(dd)=50+50-5,35=94,65(g)`

`C5_(KCl)=\frac{0,15.74,5}{94,65}.100=11,8%`

`3,`

`n_(Cu(OH)_2)=\frac{11,76}{98}=0,12(mol)`

$Cu(OH)_2\xrightarrow{t^o}CuO+H_2O$

`0,12`                        `0,12`

`CuO+H_2SO_4->CuSO_4+H_2O`

`0,12`          `0,12`          `0,12`

`a,`

`m_(CuO)=0,12.80=9,6(g)`

`b,`

`C_(MH_2SO_4)=\frac{0,12}{0,1}=1,2M`

`c,`

`C_(MCuSO_4)=\frac{0,12}{0,1}=1,2M`

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
AgN{O_3} + HCl \to AgCl + HN{O_3}\\
{m_{AgN{O_3}}} = \dfrac{{50 \times 17\% }}{{100\% }} = 8,5g\\
 \to {n_{AgN{O_3}}} = 0,05mol\\
a)\\
{n_{AgCl}} = {n_{AgN{O_3}}} = 0,05mol\\
 \to {m_{AgCl}} = 7,2g\\
b)\\
{n_{HCl}} = {n_{AgN{O_3}}} = 0,05mol \to {m_{HCl}} = 1,825g\\
 \to C{\% _{HCl}} = \dfrac{{1,825}}{{20}} \times 100\%  = 9,125\% \\
c)\\
{n_{HN{O_3}}} = {n_{AgN{O_3}}} = 0,05mol \to {m_{HN{O_3}}} = 3,15g\\
 \to {m_{{\rm{dd}}}} = {m_{{\rm{dd}}AgN{O_3}}} + {m_{{\rm{dd}}HCl}} - {m_{AgCl}} = 50 + 20 - 7,2 = 62,8g\\
 \to C{\% _{HN{O_3}}} = \dfrac{{3,15}}{{62,8}} \times 100\%  = 5,02\% \\
2.\\
FeC{l_3} + 3KOH \to Fe{(OH)_3} + 3KCl\\
a)\\
{m_{FeC{l_3}}} = \dfrac{{50 \times 16,25\% }}{{100\% }} = 8,125g\\
 \to {n_{FeC{l_3}}} = 0,05mol\\
 \to {n_{Fe{{(OH)}_3}}} = {n_{FeC{l_3}}} = 0,05mol \to {m_{Fe{{(OH)}_3}}} = 5,35g\\
b)\\
{n_{KOH}} = 3{n_{FeC{l_3}}} = 0,15mol \to {m_{KOH}} = 8,4g\\
 \to C{\% _{{\rm{dd}}KOH}} = \dfrac{{8,4}}{{50}} \times 100\%  = 16,8\% \\
c)\\
{n_{KCl}} = 3{n_{FeC{l_3}}} = 0,15mol \to {m_{KCl}} = 11,175g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}FeC{l_3}}} + {m_{{\rm{dd}}KOH}} - {m_{Fe{{(OH)}_3}}} = 94,65g\\
 \to C{\% _{KCl}} = \dfrac{{11,175}}{{94,65}} \times 100\%  = 11,8\% \\
3.\\
Cu{(OH)_2} \to CuO + {H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
a)\\
{n_{Cu{{(OH)}_2}}} = 0,12mol\\
 \to {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,12mol \to {m_{CuO}} = 9,6g\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{CuO}} = 0,12mol\\
 \to C{M_{{H_2}S{O_4}}} = \dfrac{{0,12}}{{0,1}} = 1,2M\\
c)\\
{n_{{\rm{CuS}}{O_4}}} = {n_{CuO}} = 0,12mol\\
 \to C{M_{{\rm{CuS}}{O_4}}} = \dfrac{{0,12}}{{0,1}} = 1,2M
\end{array}\)