Em tham khảo nha:
\(\begin{array}{l}
*\\
{n_{Al}} = \dfrac{{{m_{Al}}}}{{{M_{Al}}}} = \dfrac{{0,27}}{{27}} = 0,01\,mol\\
*\\
{n_{NaCl}} = \dfrac{{{m_{NaCl}}}}{{{M_{NaCl}}}} = \dfrac{{11,7}}{{58,5}} = 0,2\,mol\\
*\\
{n_{CuS{O_4}.5{H_2}O}} = \dfrac{{{m_{CuS{O_4}.5{H_2}O}}}}{{{M_{CuS{O_4}.5{H_2}O}}}} = \dfrac{{12,5}}{{250}} = 0,05\,mol\\
*\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
*\\
{n_{{H_2}S{O_4}}} = {C_M} \times V = 0,2 \times 0,6 = 0,12\,mol\\
*\\
{n_{hh}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
*\\
{n_{HCl}} = {C_M} \times V = 0,3 \times 0,3 = 0,09\,mol\\
{n_{{H_2}S{O_4}}} = {C_M} \times V = 0,3 \times 0,9 = 0,27\,mol\\
*\\
{m_{MgC{l_2}}} = {m_{{\rm{dd}}}} \times {C_\% } = 200 \times 19\% = 38g\\
{n_{MgC{l_2}}} = \dfrac{{{m_{MgC{l_2}}}}}{{{M_{MgC{l_2}}}}} = \dfrac{{38}}{{95}} = 0,4\,mol
\end{array}\)
