Đáp án đúng: D
246,32.
nFe3+ = 0,24 mol; nAl3+ = 0,32 mol; nH+ = 0,8 mol; nSO42- = 0,88 mol.
nBa2+ = 1,3 mol, nOH- = 2,6 (mol)
$\begin{array}{l}B{{a}^{2+}}\,\,+\,\,SO_{4}^{2-}\,\xrightarrow{{}}\,BaS{{O}_{4}}\downarrow \\0,88\,\leftarrow \,0,88\,\,\,\,\to \,\,\,\,\,\,\,\,\,0,88\end{array}$
$\begin{array}{l}\,{{H}^{+}}\,+\,\,O{{H}^{-}}\,\xrightarrow{{}}\,{{H}_{2}}O\\0,8\,\to \,0,8\end{array}$
$\begin{array}{l}\,F{{e}^{3+}}\,\,+\,\,3O{{H}^{-}}\,\xrightarrow{{}}\,Fe{{(OH)}_{3}}\downarrow \\0,24\,\to \,\,0,72\,\,\,\,\,\,\,\xrightarrow{{}}\,0,24\end{array}$
$\begin{array}{l}\,\,A{{l}^{3+}}\,+\,\,3O{{H}^{-}}\,\xrightarrow{{}}\,Al{{(OH)}_{3}}\downarrow \\0,32\,\to \,0,96\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,0,32\end{array}$
$\begin{array}{l}Al{{(OH)}_{3}}\,+\,O{{H}^{-}}\,\xrightarrow{{}}\,Al(OH)_{4}^{-}\\\,\,\,0,12\,\,\,\leftarrow \,\,0,12\end{array}$
m↓ = mFe(OH)3 + mAl(OH)3 + mBaSO4 = 0,24.107 + 0,2.78 + 0,88.233 = 246,32 (gam)
