`n_{H_2}=n_{Zn}=\frac{7,84}{22,4}=0,35(mol)`

`=> %m_{Zn}=\frac{0,35.65.100%}{25,99}\approx 85,53%`

`m_{ZnO}=25,99-0,35.65=3,24g`

`=> %m_{ZnO}=\frac{3,24.100%}{25,99}\approx 12,5%`

`Zn+H_2SO_4\to ZnSO_4+H_2`

`ZnO+H_2SO_4\to ZnSO_4+H_2O`

`b)``n_{HCl}=n_{Zn}+n_{ZnO}=0,35+\frac{3,24}{81}=0,39(mol)`

`C%_{H_2SO_4}=\frac{0,39.98}{250}.100%=15,3%`

`c)` BTKL: `m_{\text{dd spu}}=m_{\text{hh}}+m_{H_2SO_4}-m_{H_2}`

`=> m_{\text{dd spu}}=25,99+250-0,35.2=275,3%`

`n_{ZnSO_4}=n_{Zn}+n_{ZnO}=0,39(mol)`

`=> C%_{ZnSO_4}=\frac{161.0,39.100%}{275,3}\approx 22,8%`

`a)`

`ZnO+ H_2SO_4→ ZnSO_4+ H_2O` 

`Zn + H_2SO_4→ ZnSO_4+ H_2` 

`n_{H_2} = {7,84}/{22,4} = 0,35(mol)`

`n_{Zn} = n_{H_2} = 0,35(mol)` 

`%m_{Zn} ={0,35.  65}/{25,99} .100= 87,53%`

`%m_{ZnO} = 100-87,53=12,47%`

`b)`

`n_{ZnO}={25,99-0,35.65}/{81} = 0,04(mol)`

`n_{H_2SO_4} = n_{Zn} + m_{ZnO}`

`= 0,35+ 0,04= 0,39(mol)`

`C%_{H_2SO_4} = {0,39. 98}/{250} .100= 15,288%`

`c)`

`mdd= 25,99-+250-0,35.2= 275,29(g)`

`n_{ZnSO_4} = n_{H_2SO_4} = 0,39(mol)`

`C%_{ZnSO_4}]={0,39. 161}/{275,29} .100=22,8%`