Đáp án:

 b) \({\text{\% }}{{\text{m}}_{MgO}} = 52,6\%  \to \% {m_{Al}} = 47,4\% \)

\({m_{muối}} = 11,04{\text{ gam}}\)

Giải thích các bước giải:

 Gọi số mol MgO là x, Al là y.

Phản ứng xảy ra:

\(MgO + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}O\)

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

Ta có:  \({m_{{H_2}S{O_4}}} = 73,5.12\%  = 8,82{\text{ gam}} \to {{\text{n}}_{{H_2}S{O_4}}} = \frac{{8,82}}{{98}} = 0,09{\text{ mol = }}{{\text{n}}_{MgO}} + \frac{3}{2}{n_{Al}} = x + 1,5y\)

\({n_{{H_2}}} = \frac{{1,344}}{{22,4}} = 0,06{\text{ mol = }}\frac{3}{2}{n_{Al}} = 1,5y \to y = 0,04 \to x = 0,03\)

\(\to {m_{MgO}} = 0,03.40 = 1,2{\text{ gam; }}{{\text{m}}_{Al}} = 0,04.27 = 1,08{\text{ gam}} \to {\text{\% }}{{\text{m}}_{MgO}} = \frac{{1,2}}{{1,2 + 1,08}} = 52,6\%  \to \% {m_{Al}} = 47,4\% \)

BTKL: \({m_{muối}} = {m_X} + {m_{{H_2}S{O_4}}} - {m_{{H_2}}} = 1,2 + 1,08 + 8,82 - 0,03.2 = 11,04{\text{ gam}}\)