a) Al2O3 + 3H2SO4 ➜ Al2(SO4)3 + 3H2O (1)
MgO + H2SO4 ➜ MgSO4 + H2O (2)
b) \(m_{H_2SO_4}=400\times9,8\%=39,2\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
Gọi số mol lần lượt của Al2O3 và MgO là x,y
Theo PT1: \(n_{H_2SO_4}=3n_{Al_2O_3}=3x\left(mol\right)\)
Theo PT2: \(n_{H_2SO_4}=n_{MgO}=y\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}3x+y=0,4\\102x+40y=14,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
Vậy \(n_{Al_2O_3}=0,1mol\) , \(n_{MgO}=0,1mol\)
c) Theo PT1: \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1\times342=34,2\left(g\right)\)
Theo PT2: \(n_{MgSO_4}=n_{MgO}=0,1mol\)
\(\Rightarrow m_{MgSO_4}=0,1\times120=12\left(g\right)\)
\(m_{dd}=14,2+400=414,2\left(g\right)\)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{34,2}{414,2}\times100\%\approx8,26\%\)
\(C\%_{ddMgSO_4}=\dfrac{12}{414,2}\times100\%\approx2,9\%\)
