$n_{H_2}= \frac{1,12}{22,4}= 0,05 mol$
$Zn+ 2HCl \rightarrow ZnCl_2+ H_2$
=> $n_{Zn}= 0,05 mol; n_{HCl}= 0,1 mol$
$m_{Zn}= 0,05.65= 3,25g$
$m_{HCl}= 0,1.36,5= 3,65g$
$2H_2+ O_2 \buildrel{{t^o}}\over\longrightarrow 2H_2O$
=> $n_{O_2}= 0,025 mol$
=> $V_{kk}= 5V_{O_2}= 5.0,025.22,4= 2,8l$