Answer
See the answer below.
Work Step by Step
The number of moles of $CO_2$ is:
$n=(745/760×1\ atm)×0.665\ L/((0.082)(298\ K))$
$n=0.0267\ mol$
From stoichiometry, that's the total number of moles of both compounds.
Let x be the number of moles of $NaHCO_3$ in the sample:
$x×84.0\ g/mol+(0.0267-x)×106.0\ g/mol=2.50\ g$
Solving for x:
$x=0.0150\ mol\ NaHCO_3$
Therefore, there are: $0.0117\ mol\ Na_2CO_3$
Weight percentages:
$0.0150\ mol×84\ g/mol/2.50\ g×100\%=50.4\%\ NaHCO_3$
$0.0117\ mol×106\ g/mol/2.50\ g×100\%=49.6\%\ Na_2CO_3$
