Đáp án:
a=54,26
Giải thích các bước giải:
\(\begin{array}{l}
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
N{a_2}O + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{41,16}}{{98}} = 0,42\,mol\\
Theo\,pt:{n_{{H_2}O}} = {n_{{H_2}S{O_4}}} = 0,42\,mol\\
BTKL:\\
{m_{hh}} + {m_{{H_2}S{O_4}}} = m + {m_{{H_2}O}}\\
\Rightarrow m = 20,66 + 0,42 \times 98 - 0,42 \times 18 = 54,26g
\end{array}\)
