a)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
b)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{196.20\%}{98}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\) => Fe2O3 hết, H2SO4 dư
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
0,1------->0,3--------->0,1
\(m_{Fe_2\left(SO_4\right)_3}=0,1.400=40\left(g\right)\)
c)
mdd sau pư = 16 + 196 = 212 (g)
\(\left\{{}\begin{matrix}C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{40}{212}.100\%=18,868\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,3\right).98}{212}.100\%=4,623\%\end{matrix}\right.\)