Answer 1

\(Cu^{2+}+2OH^-\rightarrow Cu\left(OH\right)_2\\ Fe^{3+}+3OH^-\rightarrow Fe\left(OH\right)_3\)

\(n_{Cu^{2+}}=0,02\left(mol\right);n_{Fe^{3+}}=0,015\left(mol\right)\\ n_{OH^-}=0,035.2+0,02=0,09>2n_{Cu^{2+}}+3n_{Fe^{3+}}\)

Do đó $OH^-$ dư

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\\ n_{Ba^{2+}}=0,035>n_{SO_4^{2-}}=0,02\)

Do đó $Ba^{2+}$ dư

Suy ra : 

$m_{kết\ tủa} = 0,02.98 + 0,015.107 + 0,02.233 = 8,225(gam)$