Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
C{\% _{Zn{{(N{O_3})}_2}}} = \dfrac{{18,9}}{{18,9 + 281,1}} \times 100\%  = 6,3\% \\
b)\\
C{\% _{AgN{O_3}}} = \dfrac{{34}}{{200}} \times 100\%  = 17\% \\
2)\\
a)\\
{m_{{\rm{dd}}}} = \dfrac{{8 \times 100}}{{10}} = 80g\\
b)\\
{m_{dd}} = \dfrac{{16 \times 100}}{{20}} = 80g\\
3)\\
a)\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 9,8}}{{100}} = 19,6g\\
{n_{{H_2}S{O_4}}} = \dfrac{{19,6}}{{98}} = 0,2mol\\
c)\\
{m_{NaOH}} = \dfrac{{300 \times 10}}{{100}} = 30g\\
{n_{NaOH}} = \dfrac{{30}}{{40}} = 0,75mol\\
b)\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{150 \times 20}}{{100}} = 30g\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{30}}{{400}} = 0,075mol\\
d)\\
{m_{MgC{l_2}}} = \dfrac{{400 \times 9,5}}{{100}} = 38g\\
{n_{MgC{l_2}}} = \dfrac{{38}}{{95}} = 0,4mol\\
4)\\
a)\\
{n_{KOH}} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{C_{{M_{KOH}}}} = \dfrac{{0,2}}{2} = 0,1M\\
b)\\
{n_{AlC{l_3}}} = \dfrac{{2,67}}{{133,5}} = 0,02mol\\
{C_{{M_{AlC{l_3}}}}} = \dfrac{{0,02}}{{0,1}} = 0,2M\\
5)\\
a)\\
{n_{Ca{{(OH)}_2}}} = \dfrac{{14,8}}{{74}} = 0,2mol\\
{V_{Ca{{(OH)}_2}}} = \dfrac{{0,2}}{{0,2}} = 1l\\
b)\\
{n_{Pb{{(N{O_3})}_2}}} = \dfrac{{6,62}}{{331}} = 0,02mol\\
{V_{Pb{{(N{O_3})}_2}}} = \dfrac{{0,02}}{{0,1}} = 0,2l = 200ml
\end{array}\)

Lời giải chi tiết `:`

 Bài `1:`

$a/$

`m_{dd}=m_{Zn(NO_3)_2}+m_{H_2O}=18,9+281,1=300(g)`

`->` `%_{Zn(No_3)_2}={18,9}/300 .100%=0,063.100%=6,3%`

$b/$

`C%_{AgNo_3}={m_{AgNO_3}}/{m_{dd}}=34/200 .100%=0,17.100%=17%`

`----------`

Bài `2:`

$a/$
`m_{dd CuSO_4}={mct.100}/C={8.100}/10=800/10=80(g)`

$b/$

`m_{dd BaCl_2}={mct.100}/C={16.100}/20=1600/20=80(g)`

`----------` 

Bài `3:`

$a/$

`200` gam dung dịch `H_2SO_4`  `9,8%`

`->` `m_{H_2SO_4} = { 200. 9,8}/ 100=1960/100=19,6 (g)`

`->` `n_{H_2SO_4} = {19,6}/98=0,2 (mol)`

$b/$ 

`150` gam dung dịch `Fe_2(SO_4)_3` `20%`

`->` `m_{Fe_2(SO_4)_3}=(150.20)/100=30(g)`

`->` `n_{Fe_2(SO_4)_3}=30/400=0,075(mol)`

$c/$

`300` gam dung dịch `NaOH`    `10%`

`->` `m_{NaOH}=(300.10)/100=3000/100=30(g)`

`->` `n_{NaOH}=30/40=0,75(mol)`

$d/$

`400` gam dung dịch `MgCl_2`     `9,5 %`

`->` `m_(MgCl_2)=(400.9,5)/100=3800/100=38(g)`

`->` `n_(MgCI_2)= 38/95=0,4(mol)`

`----------`

Bài `4:`

$a/$

`n_(KOH)={11,2}/(56)=0,2(mol)`

`C_(M_(KOH))=(0,2)/2=0,1M`

$b/$

`n_(AICI_3)=(2,67)/(133,5)=0,02(mol)`

`C_(m_(AICI_3))=(0,02)/(0,1)=0,2M`

`----------` 

Bài `5:`

$a/$

`n_{Ca(OH)_2}=(14,8)/74=0,2(mol)`

`V_{Ca(OH)_2}=(0,2)/(0,2)=1l`

$b/$

`n_{pb(NO_3)_2}=(6,62)/331=0,02(mol)`

`V_{pb(NO_3)_2}=(0,02)/(0,1)=0,2l=200ml`