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2)

Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

Ta có:

\({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol;}}{{\text{n}}_{HCl}} = \frac{{8,76}}{{36,5}} = 0,24{\text{ mol}}\)

\( \to {n_{HCl}} > 2{n_{Zn}}\) nên \(HCl\) dư

\( \to {n_{HCl{\text{ dư}}}} = 0,24 - 0,1.2 = 0,04{\text{ mol}}\)

\({n_{{H_2}}} = {n_{Zn}} = 0,1{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

\({n_{ZnC{l_2}}} = {n_{Zn}} = 0,1{\text{ mol}}\)

\( \to {m_{ZnC{l_2}}} = 0,1.(65 + 35,5.2) = 13,6{\text{ gam}}\)

3)

Phản ứng xảy ra:

\(4{\text{A}}l + 3{{\text{O}}_2}\xrightarrow{{{t^o}}}2{\text{A}}{l_2}{O_3}\)

Ta có:

\({n_{Al}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{3}{4}{n_{Al}} = 0,15{\text{ mol}}\)

\( \to {V_{{O_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)

\( \to {V_{kk}} = 5{V_{{O_2}}} = 3,36.5 = 16,8{\text{ lít}}\)

Điều chế khí \(O_2\)

\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

\( \to {n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,15.2 = 0,3{\text{ mol}}\)

\( \to {m_{KMn{O_4}}} = 0,3.(39 + 55 + 16.4) = 47,4{\text{ gam}}\)