Cho $\begin{cases} Na_2SO_4:x(mol)\\ K_2SO_4: 2x(mol)\\\end{cases}$
`=> m_{\text{dd A}}=102+m_{Na_2SO_4}+m_{K_2SO_4}`
`=> m_{\text{dd A}}=102+490x(g)`
`n_{BaCl_2}=\frac{1664.10}{100.208}=0,8(mol)`
`n_{\downarrow}=n_{BaSO_4}=\frac{46,6}{233}=0,2(mol)`
`Na_2SO_4+BaCl_2\to BaSO_4\downarrow+2NaCl(1)`
`K_2SO_4+BaCl_2\to BaSO_4\downarrow+2KCl(2)`
Do `n_{BaCl_2}>n_{\downarrow}`
`=> n_{BaCl_2\text{dư}}`
`BaCl_2+H_2SO_4\to BaSO_4\downarrow+2HCl`
`=> n_{BaCl_2(1),(2)}=x+2x=0,8-0,2=0,6(mol)`
`=> 3x=0,6(mol)`
`=> x=0,2(mol)`
`C%_{Na_2SO_4}=\frac{0,2.142.100%}{102+490.0,2}=14,2%`
`C%_{K_2SO_4}=\frac{0,4.174.100%}{102+490.0,2}=34,8%`