Đáp án:
b) \({{\text{V}}_{{O_2}}} = 4,48{\text{ lít;}}{{\text{V}}_{kk}} = 22,4{\text{ lít}}\)
c) \({{\text{m}}_{A{l_2}{O_3}}} = 10,2{\text{ gam;}}{{\text{m}}_{MgO}} = 4{\text{ gam}}\)
d) \({m_{KMn{O_4}}} = 79{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({m_{Al}} = \frac{{7,8}}{{4 + 9}}.9 = 5,4{\text{ gam}} \to {{\text{m}}_{Mg}} = 7,8 - 5,4 = 2,4{\text{ gam}}\)
\( \to {n_{Al}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol;}}{{\text{n}}_{Mg}} = \frac{{2,4}}{{24}} = 0,1{\text{ mol}}\)
Phản ứng xảy ra:
\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)
\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)
Ta có:
\({n_{{O_2}}} = \frac{1}{2}{n_{Mg}} + \frac{3}{4}{n_{Al}} = \frac{1}{2}.0,1 + \frac{3}{4}.0,2 = 0,2{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 0,2.22,4 = 4,48{\text{ lít}} \to {{\text{V}}_{kk}} = 4,48.5 = 22,4{\text{ lít}}\)
\({n_{A{l_2}{O_3}}} = \frac{1}{2}{n_{Al}} = 0,1{\text{ mol;}}{{\text{n}}_{MgO}} = {n_{Mg}} = 0,1{\text{ mol}} \to {{\text{m}}_{A{l_2}{O_3}}} = 0,1.(27.2 + 16.3) = 10,2{\text{ gam;}}{{\text{m}}_{MgO}} = 0,1.(24 + 16) = 4{\text{ gam}}\)
\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\( \to {n_{KMn{O_4}{\text{ lý thuyết}}}} = 2{n_{{O_2}}} = 0,4{\text{ mol}} \to {{\text{n}}_{KMn{O_4}{\text{ thực tế}}}} = \frac{{0,4}}{{80\% }} = 0,5 \to {m_{KMn{O_4}}} = 0,5.158 = 79{\text{ gam}}\)
