\(\begin{array}{l}
a)\\
n_{HNO_3}=\frac{160.6,3\%}{63}=0,16(mol)\\
ZnO+2HNO_3\to Zn(NO_3)_2+H_2O\\
Fe_2O_3+6HNO_3\to 2Fe(NO_3)_3+3H_2O\\
Theo\,PT:\,2n_{ZnO}+6n_{Fe_2O_3}=n_{HNO_3}=0,16(mol)\\
\to n_{ZnO}=n_{Fe_2O_3}=0,02(mol)\\
\to x=0,02(81+160)=4,82(g)\\
b)\\
m_{dd\,spu}=4,82+160=164,82(g)\\
Theo\,PT:\\
n_{Zn(NO_3)_2}=n_{ZnO}=0,02(mol)\\
n_{Fe(NO_3)_3}=2n_{Fe_2O_3}=0,04(mol)\\
\to \begin{cases} C\%_{Zn(NO_3)_2}=\frac{0,02.189}{164,82}.100\%\approx 2,29\%\\ C\%_{Fe(NO_3)_3}=\frac{0,04.242}{164,82}.100\%\approx 5,87\%\end{cases}
\end{array}\)

Đáp án:

`a.\ x=4,82(g)`

`b.\ {(C%_(Zn(NO_3)_2)≈2text(,)29%),(C%_(Fe(NO_3)_3)≈5text(,)87%):}`

Giải thích các bước giải:

Gọi `n_(ZnO)=n_(Fe_2O_3)=a(mol)`

`m_(HNO_3)=160xx6,3%=10,08(g)`

`=>n_(HNO_3)=(10,08)/(63)=0,16(mol)`

`ZnO+2HNO_3->Zn(NO_3)_2+H_2O`

`Fe_2O_3+6HNO_3->2Fe(NO_3)_3+H_2O`

`∑n_(HNO_3)=2n_(ZnO)+6n_(Fe_2O_3)=2a+6a=8a=0,16(mol)`

`<=>a=0,02`

`=>n_(ZnO)=n_(Fe_2O_3)=0,02(mol)`

`=>x=81xx0,02+160xx0,02=4,82(g)`

`b.`

`n_(Zn(NO_3)_2)=n_(ZnO)=0,02(mol)`

`n_(Fe(NO_3)_3)=2n_(Fe_2O_3)=0,04(mol)`

`m_(text(ddspư))=160+4,82=164,82(g)`

`C%_(Zn(NO_3)_2)=(0,02xx189)/(164,82)xx100≈2,29%`

`C%_(Fe(NO_3)_3)=(0,04xx242)/(164,82)xx100≈5,87%`