`n_{Fe_2O_3}=16/160=0.1\ (mol)`
`n_{H_2SO_4}=\frac{300.20}{98.100}=30/49\ (mol)`
PTHH:
`Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O`
Ta có tỉ lệ:
$\dfrac{n_{Fe_2O_3}}{1}=\dfrac{0,1}{1}=0,1<\dfrac{n_{H_2SO_4}}3=\dfrac{\dfrac{30}{49}}{3}=0,2$
`\to Fe` hết, `H_2SO_4` dư
Theo PT: $\left\{\begin{matrix}n_{Fe_2(SO_4)_3}=n_{Fe_2O_3}=0,1\ (mol)\\n_{H_2SO_4\ pứ}=3n_{Fe_2O_3}=3.0,1=0,3\ (mol)\end{matrix}\right.$
`\to n_{H_2SO_4\ dư}=30/49-0,3=0,312\ (mol)`
$m_{dd\ spứ}=16+300=316\ (g)$
`\to`$\left\{\begin{matrix}{C\%}_{dd\ Fe_2(SO_4)_3}=\dfrac{0,1.400}{316}.100\%=12,66\%\\{C\%}_{dd\ H_2SO_4\ dư}=\dfrac{0,312.98}{316}.100\%=9,67\%\end{matrix}\right.$
