Giải thích các bước giải:
\(\begin{array}{l}
2.\\
2Al + 3C{\rm{uS}}{O_4} \to A{l_2}{(S{O_4})_3} + 3Cu\\
{n_{{\rm{CuS}}{O_4}}} = 0,36mol\\
{n_{Al}} = 0,2mol\\
\to {n_{Al}} < {n_{{\rm{CuS}}{O_4}}} \to {n_{{\rm{CuS}}{O_4}}}du\\
\to {n_{Cu}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
\to m = {m_{Cu}} = 19,2g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,1mol\\
\to C{M_{A{l_2}{{(S{O_4})}_3}}} = \frac{{0,1}}{{0,2}} = 0,5M\\
{n_{{\rm{CuS}}{O_4}(pt)}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
\to {n_{{\rm{CuS}}{O_4}(dư)}} = 0,06mol\\
\to C{M_{{\rm{CuS}}{O_4}(dư)}} = \dfrac{{0,06}}{{0,2}} = 0,3M\\
4.\\
Fe + S \to FeS\\
FeS + 2HCl \to FeC{l_2} + {H_2}S\\
a)\\
{n_{Fe}} = 0,03mol\\
{n_{FeS}} = 0,03mol\\
\to {n_S} = {n_{Fe}} = 0,03mol\\
\to {m_S} = 0,96g\\
b)\\
{n_{HCl}} = 2{n_{FeS}} = 0,06mol\\
\to {V_{HCl}} = \dfrac{{0,06}}{2} = 0,03l\\
5.\\
Cu + C{l_2} \to CuC{l_2}\\
{n_{Cu}} = 0,1mol\\
{n_{CuC{l_2}}} = 0,1mol\\
\to {n_{C{l_2}}} = {n_{Cu}} = 0,1mol\\
\to {V_{C{l_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)
