`Zn+2HCl->ZnCl_2+H_2`
`n_{H_2}=n_{Zn}=(9,75)/(65)=0,15mol`
`V_{H_2}=0,15.22,4=3,36l`
`4H_2+Fe_3O_4->3Fe+4H_2O`
`n_{H_2}=(0,15)/4=0,0375mol<n_{Fe_3O_4}=(23,2)/(56.3+16.4)=0,1mol`
`n_{Fe}=3/4.0,15=0,1125mol`
`n_{Fe_3O_4(pu)}=1/4 .0,15=0,0375mol`
`n_{Fe_3O_4(du)}=0,1-0,0375=0,0625mol`
`m_{CR}=m_{Fe}+m_{Fe_3O_4}=0,1125.56+0,0625.(56.3+16.4)=20,8g`