\(1.\)

\(n_{H_2^{}}^{} = \frac{{8,96}}{{22,4}} = 0,4\;{\text{mol}}\)

\(Al:x\;{\text{mol; Fe: y mol}}\)

\(2Al + 3H_2^{}SO_4^{} \to Al_2^{}(SO_4^{})_3^{} + 3H_2^{}\)

\({\text{ x          }}\frac{3}{x}{\text{x                }}\frac{x}{2}\;{\text{           }}\frac{3}{2}x\)

\(Fe + H_2^{}SO_4^{} \to FeSO_4^{} + H_2^{}\)

\(\,{\text{y       y                    y       y }}\)

\(27x + 56y = 11\)

\(\frac{3}{2}x + y = 0,4\)

\( \to x = 0,2;y = 0,1\)

\(\% m_{Al}^{} = \frac{{0,2.27}}{{11}}.100 = 49\% \)

\(\% m_{Fe}^{} = 100 - 49 = 51\% \)

\(b)m_{H_2^{}SO_4^{}}^{} = 98.(0,2.\frac{3}{2} + 0,1) = 39,2g\)

\(c)m_{Muoi}^{} = \frac{1}{2}.0,2.342 + 0,1.152 = 49,4g\)

\(2.n_{Zn}^{} = \frac{{19,5}}{{65}} = 0,3\;{\text{mol}}\)

\(n_{H_2^{}SO_4^{}}^{} = \frac{{39,2}}{{98}} = 0,4\;{\text{mol}}\)

\(Zn + H_2^{}SO_4^{} \to ZnSO_4^{} + H_2^{}\)

\({\text{ }}\frac{{0,3}}{1} < \frac{{0,4}}{1} \to tinh{\text{ theo Zn}}\)

\(0,3{\text{                                  0}}{\text{,3}}\)

\(b)V_{H_2^{}}^{} = 0,3.22,4 = 6,72l\)

\(c)CuO + H_2^{} \to Cu + H_2^{}O\)

\(PbO + H_2^{} \to Pb + H_2^{}O\)

\({\text{1 mol H}}_2^{}{\text{ se lay 1 mol O de tao ra 1 mol H}}_2^{}O\)

\( \to m_{Giam}^{} = m_{O_{BiLayDi}^{}}^{} = n_{H_2^{}}^{}.16 = 0,3.16 = 4,8g\)

\(3.Ca(OH)_2^{} + CO_2^{} \to CaCO_3^{} + H_2^{}O\)

\({\text{                         0}}{\text{,5       0}}{\text{,5}}\)

\(CaCO_3^{}\xrightarrow{{{t^o}}}CaO + CO_2^{}\)

\({\text{    0}}{\text{,5                0}}{\text{,5}}\)

\(CH_4^{} + 2O_2^{}\xrightarrow{{{t^o}}}CO_2^{} + 2H_2^{}O\)

\({\text{  x                          x}}\)

\(C_2^{}H_4^{} + 3O_2^{}\xrightarrow{{{t^o}}}2CO_2^{} + 2H_2^{}O\)

\({\text{  y                            2y}}\)

\(n_{CaO}^{} = \frac{{28}}{{56}} = 0,5\;{\text{mol}}\)

\( \to n_{CO_2^{}}^{} = 0,5{\text{ mol}}\)

\({\text{CH}}_4^{}:x;C_2^{}H_4^{}:y\)

\(16x + 28y = 7,6\)

\(x + 2y = 0,5\)

\( \to x = 0,3;y = 0,1\)

\(\% m_{CH_4^{}}^{} = \frac{{0,3.16}}{{7,6}}.100 = 63,15\% \)

\(\% m_{C_2^{}H_4^{}}^{} = 100 - 63,15 = 36,85\% \)

\(\% V_{CH_4^{}}^{} = \frac{{0,3}}{{0,4}}.100 = 75\% \)

\(\% V_{C_2^{}H_4^{}}^{} = \frac{{0,1}}{{0,4}}.100 = 25\% \)