7,2 gam chất rắn không tan là Cu dư `=> m_{Cu(dư)} = 7,2(g)`
Đặt \(n_{Fe_3O_4}=x\left(mol\right)\)
PTHH:
\(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)
x-------------------->x----------->2x
\(2FeCl_3+Cu\rightarrow2FeCl_2+CuCl_2\)
2x-------------------------------->x
BTNT Fe: \(n_{Fe_2O_3}=\dfrac{3}{2}n_{Fe_3O_4}=1,5x\left(mol\right)\)
BTNT Cu: \(n_{Cu}=n_{CuO}=n_{CuCl_2}=x\left(mol\right)\)
`=> 80x + 160.1,5x = 48`
`=> x = 0,15(mol)`
`=> m = 0,15.232 + 0,15.64 + 7,2 = 51,6(g)`
Theo PT: `n_{HCl} = 8n_{Fe_3O_4} = 1,2(mol)`
`=> a = C\%_{HCl} = (1,2.36,5)/(200).100\% = 21,9\%`