Dung dịch A có nồng độ \(\left\{{}\begin{matrix}H_2SO_4:aM\\FeSO_4:bM\\XSO_4:cM\end{matrix}\right.\)
- TN1:
\(n_{H_2SO_4}=0,2a\left(mol\right)\)
\(n_{NaOH}=0,04.0,5=0,02\left(mol\right)\)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,02---->0,01
=> 0,2.a = 0,01
=> a = 0,05
- TN2:
\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,01\left(mol\right)\\n_{FeSO_4}=0,2b\left(mol\right)\\n_{XSO_4}=0,2c\left(mol\right)\end{matrix}\right.\)
\(n_{NaOH}=0,3.0,5=0,15\left(mol\right)\)
\(n_{HCl}=0,04.0,25=0,01\left(mol\right)\)
\(n_{H^+}=0,03\left(mol\right);n_{OH^-}=0,15\left(mol\right);n_{Fe^{2+}}=0,2b\left(mol\right);n_{X^{2+}}=0,2c\left(mol\right)\)
\(n_{SO_4^{2-}}=0,01+0,2b+0,2c\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
0,03->0,03
\(Fe^{2+}+2OH^-\rightarrow Fe\left(OH\right)_2\)
0,2b-->0,4b------->0,2b
\(X^{2+}+2OH^-\rightarrow X\left(OH\right)_2\)
0,2c-->0,4c------>0,2c
=> \(n_{OH^-}=0,4b+0,4c+0,03=0,15\left(mol\right)\)
=> b + c = 0,3
Kết tủa gồm \(\left\{{}\begin{matrix}BaSO_4:0,01+0,2b+0,2c\left(mol\right)\\Fe\left(OH\right)_2:0,2b\left(mol\right)\\X\left(OH\right)_2:0,2c\left(mol\right)\end{matrix}\right.\)
=> \(233\left(0,01+0,2b+0,2c\right)+18b+0,2.c.M_X+6,8c=21,07\)
=> 0,2.c.MX + 64,6b + 53,4c = 18,74
=> 0,2.c.MX - 11,2c = -0,64
=> \(c\left(0,2.M_X-11,2\right)=-0,64\)
Mà 0 < c < 0,3
=> \(M_X< 45,33\left(g/mol\right)\)
=> X là Mg