Đáp án:
\({m_{dd\;{{\text{H}}_2}S{O_4}}} = 189,7{\text{ gam}}\)
\(R\) là \(Mg\)
\( a\% = 15\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
\(R + {H_2}S{O_4}\xrightarrow{{}}RS{O_4} + {H_2}\)
Ta có:
\({m_{dd\;{{\text{H}}_2}S{O_4}}} = 169,375.1,12 = 189,7{\text{ gam}}\)
\({n_{{H_2}}} = \frac{{12,32}}{{22,4}} = 0,55{\text{ mol}}\)
BTKL:
\({m_{hh}} + {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = {m_{dd\;{\text{B}}}} + {m_{{H_2}}}\)
\( \to 11,4 + 189,7 = {m_{dd\;{\text{B}}}} + 0,55.2\)
\( \to {m_{dd\;{\text{B}}}} = 200{\text{ gam}}\)
\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = 200.17,1\% = 34,2{\text{ gam}}\)
\( \to {n_{A{l_2}{{(S{O_4})}_3}}} = \frac{{34,2}}{{27.2 + 96.3}} = 0,1{\text{ mol}}\)
\( \to {n_{Al}} = 2{n_{A{l_2}{{(S{O_4})}_3}}} = 0,1.2 = 0,2{\text{ mol}}\)
\( \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam}} \to {{\text{m}}_{Al}} = 11,4 - 5,4 = 6{\text{ gam}}\)
\({n_{{H_2}}} = \frac{3}{2}{n_{Al}} + {n_R} = \frac{3}{2}.0,2 + {n_R} = 0,55 \to {n_R} = 0,25\)
\( \to {M_R} = \frac{6}{{0,25}} = 24 \to R:Mg\)
\({n_{MgS{O_4}}} = {n_{Mg}} = 0,25{\text{ mol}}\)
\( \to {m_{MgS{O_4}}} = 0,25.120 = 30{\text{ gam}}\)
\( \to a\% = C{\% _{MgS{O_4}}} = \frac{{30}}{{200}}.100\% = 15\% \)
