Đáp án:
\({m_{{\rm{dd}}spu}} = 101,48g\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = 0,1 \times 98 = 9,8g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{9,8 \times 100}}{{10}} = 98g\\
{m_{{H_2}}} = 0,1 \times 2 = 0,2g\\
{m_{{\rm{dd}}spu}} = {m_{hh}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}}\\
{m_{{\rm{dd}}spu}} = 3,68 + 98 - 0,2 = 101,48g
\end{array}\)
