Đáp án:
6) 25,68g
7) C
Giải thích các bước giải:
\(\begin{array}{l}
6)\\
{H_2}S{O_4} + 2KOH \to {K_2}S{O_4} + 2{H_2}O\\
0,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,8\\
A{l_2}{(S{O_4})_3} + 6KOH \to 2Al{(OH)_3} + 3{K_2}S{O_4}\\
0,16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,96\,\,\,\,\,\,\,\,\,\,\,\,\,0,32\\
FeC{l_3} + 3KOH \to Fe{(OH)_3} + 3KCl\\
0,24\,\,\,\,\,\,\,\,\,\,\,0,72\,\,\,\,\,\,\,\,\,\,\,\,\,0,24\\
nKOH\,spu = 2,8 - 0,72 - 0,96 - 0,8 = 0,32\,mol\\
Al{(OH)_3} + KOH \to KAl{O_2} + 2{H_2}O\\
0,32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,32\\
\Rightarrow Al{(OH)_{3\,}}\,\tan \\
m = mFe{(OH)_3} = 0,24 \times 107 = 25,68g
\end{array}\)
