Đáp án:
\({{\text{m}}_{AgBr}} = 56,4{\text{ gam}}\)
Giải thích các bước giải:
Ta có: \({n_{Al}} = \frac{{2,7}}{{27}} = 0,1{\text{ mol}}\)
Phản ứng xảy ra:
\(2Al + 6HX\xrightarrow{{}}2Al{X_3} + 3{H_2}\)
Ta có: \({n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,15{\text{ mol}}\)
BTKL: \({m_{dd{\text{ A}}}} = {m_{Al}} + {m_{HX}} - {m_{{H_2}}} = 2,7 + 50 - 0,15.2 = 52,4{\text{ gam}}\)
\(\to {m_{Al{X_3}}} = 50,95\% .52,4 = 26,7{\text{ gam}}\)
Ta có: \({n_{Al{X_3}}} = {n_{Al}} = 0,1 \to {M_{Al{X_3}}} = 27 + 3X = \frac{{26,7}}{{0,1}} = 267 \to X = 80 \to X:Br\)
\(\to {n_{HBr}} = 3{n_{Al}} = 0,3{\text{ mol}}\)
\(AgN{O_3} + HBr\xrightarrow{{}}AgBr + HN{O_3}\)
\(\to {n_{AgBr}} = {n_{HBr}} = 0,3{\text{ mol}} \to {{\text{m}}_{AgBr}} = 0,3.(108 + 80) = 56,4{\text{ gam}}\)