Đáp án:
\(\begin{array}{l}
b)\\
{m_{Fe}} = 16,8g\\
c)\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = 16,09\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3FeS{O_4} \to 3Fe + A{l_2}{(S{O_4})_3}\\
b)\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{5,4}}{{27}} = 0,2mol\\
\Rightarrow {n_{Fe}} = \dfrac{3}{2}{n_{Al}} = \dfrac{3}{2} \times 0,2 = 0,3mol\\
{m_{Fe}} = n \times M = 0,3 \times 56 = 16,8g\\
c)\\
{m_{ddFeS{O_4}}} = V \times d = 200 \times 1,12 = 224g\\
{m_{ddspu}} = {m_{Al}} + {m_{ddFeS{O_4}}} - {m_{Fe}} = 5,4 + 224 - 16,8 = 212,6\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = n \times M = 0,1 \times 342 = 34,2g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{m_{A{l_2}{{(S{O_4})}_3}}}}}{{{m_{ddspu}}}} \times 100\% = \dfrac{{34,2}}{{212,6}} \times 100\% = 16,09\%
\end{array}\)
