\(n_{H_2SO_4}:\dfrac{171,5.20}{100.98}=0,35\left(mol\right)\)
Gọi x, y lần lượt là số mol MgO, \(Fe_2O_3\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
1..................1...................1...................(mol)
x..................x.....................x..................(mol)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
1.....................3.....................1...............(mol)
y.....................3y.....................y...............(mol)
\(\left\{{}\begin{matrix}40x+160y=16\\x+3y=0,35\end{matrix}\right.\)
->x=0,2, y= 0,05
\(m_{MgO}:0,2.40=8\left(g\right)\)
\(m_{Fe_2O_3}:16-8=8\left(g\right)\)
b)C% dd MgSO4:\(\dfrac{0,2.120}{16+171,5}.100\%=12,8\%\)
C% dd Fe2(SO4)3:\(\dfrac{0,05.400}{16+171,5}.100\%=10,67\%\)
a, Ta có nH2SO4 = \(\dfrac{171,5\times20\%}{98}\) = 0,35 ( mol )
Gọi x,y lần lượt là số mol của MgO va Fe2O3
MgO + H2SO4 → MgSO4 + H2O (1)
x → x → x → x
Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O (2)
y → 3y → y → 3y
Từ (1)(2) => \(\left\{{}\begin{matrix}40x+160y=16\\x+3y=0,35\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,2\\y=0,05\end{matrix}\right.\)
=> MMgO = 0,2 . 40 = 8 ( gam )
=> M Fe2O3 = 0,05 . 160 = 8 ( gam )
b, => MMgSO4 = 0,2 . 120 = 24 ( gam )
=> M Fe2(SO4)3 = 0,05 . 400 = 20 ( gam )
=> Mdung dịch = \(\Sigma\)Mtham gia
= MMgO + Fe2O3 + MH2SO4
= 16 +171,5 = 187,5 ( gam )
=> C%MgSO4 = \(\dfrac{24}{187,5}\) . 100 = 12,8%
=> C% Fe2(SO4)3 = \(\dfrac{20}{187,5}\) . 100 = \(\dfrac{32}{3}\)%
