a)
CTHH: M2O3
\(n_{M_2O_3}=\dfrac{4}{2.M_M+48}\left(mol\right)\)
PTHH: \(M_2O_3+3H_2SO_4\rightarrow M_2\left(SO_4\right)_3+3H_2O\)
Theo PTHH: \(n_{H_2SO_4}=3.n_{M_2O_3}=\dfrac{12}{2.M_M+48}\left(mol\right)\)
=> \(m_{H_2SO_4}=\dfrac{588}{M_M+24}\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{\dfrac{588}{M_M+24}.100}{9,8}=\dfrac{6000}{M_M+24}\left(g\right)\)
mdd sau pư = \(4+\dfrac{6000}{M_M+24}\left(g\right)\)
Theo PTHH: \(n_{M_2\left(SO_4\right)_3}=n_{M_2O_3}=\dfrac{4}{2.M_M+48}\left(mol\right)\)
Ta có: \(C\%_{M_2\left(SO_4\right)_3}=\dfrac{\dfrac{4}{2.M_M+48}\left(2.M_M+288\right)}{4+\dfrac{6000}{M_M+24}}.100\%=12,658\%\)
=> MM = 56 (g/mol)
=> M là Fe
CTHH của oxit là Fe2O3
\(m_{dd.H_2SO_4}=\) 75 (g)