## Chemistry and Chemical Reactivity (9th Edition)

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: $0.0150\ mol×84\ g/mol/2.50\ g×100\%=50.4\%\ NaHCO_3$ $0.0117\ mol×106\ g/mol/2.50\ g×100\%=49.6\%\ Na_2CO_3$