Đáp án:
b) \({{\text{m}}_{A{l_2}{O_3}}} = 2,04{\text{ gam}}\)
c,d) \({m_{{O_2}}} = 0,96{\text{ gam; }}{{\text{m}}_{kk}} = 4,35{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)
Ta có:
\({n_{Al}} = \frac{{1,08}}{{27}} = 0,04{\text{ mol}} \to {n_{A{l_2}{O_3}}} = \frac{1}{2}{n_{Al}} = 0,02{\text{ mol}} \to {{\text{m}}_{A{l_2}{O_3}}} = 0,02.(27.2 + 16.2) = 2,04{\text{ ga}}{\text{m}}\)
\({n_{{O_2}}} = \frac{3}{4}{n_{Al}} = 0,03{\text{ mol}} \to {{\text{n}}_{kk}} = 5{n_{{O_2}}} = 0,15{\text{ mol}}\)
\(\to {m_{{O_2}}} = 0,03.32 = 0,96{\text{ gam; }}{{\text{m}}_{kk}} = 0,15.29 = 4,35{\text{ gam}}\)