Giúp e vs ạ, em cảm ơn!

`a,`
Phản ứng :
$CO_{3}^{2-}+Ca^{2+}->CaCO_3$
$Mg^{2+}+2OH^{-}->Mg(OH)_2$
$Ba^{2+}+SO_{4}^{2-}->BaSO_4$
`b,`
3 tấn muối hay `3.10^{6}` (g) có :
`m_{MgCl_2}=3.10^{6}.1,224\%=36720(g)`
`m_{CaSO_4}=0,01\%.3.10^{6}=300(g)`
`m_{CaCl_2}=0,951\%.3.10^{6}=28530(g)`
`=>n_{Mg^{2+}}=n_{MgCl_2}=(36720)/(95)=387(mol)`
$n_{SO_{4}^{2-}}=n_{CaSO_4}=(300)/(136)=2,2(mol)$
$n_{Ca^{2+}}=n_{CaSO_4}+n_{CaCl_2}=(300)/(136)+(28530)/(111)=259(mol)$
`=>` Theo câu `a,:`
`n_{Na_2CO_3}=n_{Ca^{2+}}=259(mol)`
`n_{NaOH}=n_{Mg^{2+}}=387(mol)`               
`n_{BaCl_2}=2,2(mol)`
`=>m_A=259.106+387.40+208.2,2=43391,6(g)`