$_{\left[ {{H}_{3}}{{O}^{+}} \right]=\left[ HCl \right]={{10}^{-2}}mol.{{L}^{-1}}}^{\left[ {{H}_{3}}{{O}^{+}} \right]={{10}^{-PH}}={{10}^{-2}}mol.{{L}^{-1}}}$  $Zn(s)+2HCl(aq)\to ZnC{{l}_{2}}(aq)+{{H}_{2}}(g)$  $?mLCl=0/2LH2\times \frac{0/08g{{H}_{2}}}{1L{{H}_{2}}}\times \frac{1mol{{H}_{2}}}{2g{{H}_{2}}}\times \frac{2molHCl}{1mol{{H}_{2}}}\times \frac{1LHCl}{0/01molHCl}\times \frac{1000molHCl}{1LHCl}=1600mLHCl$