Answer
See the answer below.
Work Step by Step
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:
$V=0.0138\ mol×0.082×298\ K/(745/760×1\ atm)$
$V=0.344\ L$
