$\begin{align}  & NaHC{{O}_{3}}(aq)+HCl(aq)\to NaCl(aq)+{{H}_{2}}O(l)+C{{O}_{2}}(g) \\  & pH=1/15\Rightarrow \left[ {{H}^{+}} \right]={{10}^{-1/15}}={{10}^{-2}}\times {{10}^{0/85}} \\  & =7\times {{10}^{-2}}mol.{{L}^{-1}} \\  & {{M}_{HCl}}=7\times {{10}^{-2}}\frac{mol}{L} \\  & ?g\,NaHC{{O}_{3}}=0/1L\,HCl\,\,mahlol\times \frac{7\times {{10}^{-2}}mol\,HCl}{1L\,HCl} \\  & \times \frac{1mol\,NaHC{{O}_{3}}}{1mol\,HCl}\times \frac{84g\,NaHC{{O}_{3}}}{1mol\,NaHC{{O}_{3}}}=0/588g\,NaHC{{O}_{3}} \\  & ?L\,C{{O}_{2}}=7\times {{10}^{-3}}mol\,HCl\times \frac{1mol\,C{{O}_{2}}}{1mol\,HCl}\times \frac{22/4L\,C{{O}_{2}}}{1mol\,C{{O}_{2}}} \\  & =156/8\times {{10}^{-3}}L\,C{{O}_{2}} \\ \end{align}$