Đặt công thức oxit là M2O3
\(M_2O_3+3H_2SO_4\rightarrow M_2\left(SO_4\right)_3+3H_2O\)
Ta có: \(\begin{matrix}n_{M_2O_3}&=&n_{M_2\left(SO_4\right)_3}\\\dfrac{m_{M_2O_3}}{M_{M_2O_3}}&=&\dfrac{m_{M_2\left(SO_4\right)_3}}{M_{M_2\left(SO_4\right)_3}}\\\dfrac{16}{2M_M+16.3}&=&\dfrac{40}{2M_M+96.3}\\\Rightarrow M_M&=&56\left(g/mol\right)\Rightarrow\left(Fe\right)\end{matrix}\)
CT oxit: \(Fe_2O_3\)
\(n_{H_2SO_4}=3.n_{Fe_2O_3}=3.\dfrac{16}{160}=0,3\left(mol\right)\)
\(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
\(m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{C\%}=\dfrac{29,4.100\%}{10\%0\%}=294\left(g\right)\)
\(V_{ddH_2SO_4}=\dfrac{m_{ddH_2SO_4}}{D_{H_2SO_4}}=\dfrac{294}{0,69}\approx426\left(ml\right)\)
