## Chemistry and Chemical Reactivity (9th Edition)

The number of moles of Nitrogen is: $n=(713/760×1\ atm)×0.295\ L/((0.082)(294\ K))$ $n=0.0115\ mol$ From stoichiometry: $0.0115\ mol\ NaNO_2$ Which corresponds to: $0.0115\ mol×69.0\ g/mol=0.791\ g$ Mass fraction: $0.791/1.232×100\%=64.2\%$