## Chemistry and Chemical Reactivity (9th Edition)

The number of moles of HCl is: $12/1000\ L×1.50\ M=0.018\ mol$ Let x be the number of moles of $NaHCO_3$ in the sample: $x×84.0\ g/mol+(0.018-x)/2×106.0\ g/mol=1.249\ g$ Solving for x: $x=9.516×10^{-3}\ mol\ NaHCO_3$ Therefore, it follows that there are: $4.242×10^{-3}\ mol\ Na_2CO_3$ Total number of moles of $CO_2$: $0.0138\ mol$ Volume evolved: $V=0.0138\ mol×0.082×298\ K/(745/760×1\ atm)$ $V=0.344\ L$