Apparently 5.003g of a mixture of NaCl and NaHCO3 are dissolved in 65.1 L of a 1.0 M solution of HCl.The acid ,in large excess, will react rapidely with the base NaHCO3 liberating CO2 gas :
Na HCO3 + HCl = CO2 (gas) + H2O + NaCl (1)
the gas is collected and measured to be 0.802g.
NaHCO3 and CO2 in the reaction above behave as monovalent groups ( see my ansewr to Brieonna's question on neutralyzing H3PO4 with NaOH ), hence their gram equivalents are equal to their moles ( 84g and 44g respectively ). The Eq of CO2 liberated in reaction(1) = 0.802/44 =18.23 mEq.
this is also the Eq of NaHCO3 present and in grams : 18.23 X 84 = 1531 mg.
and the NaCl present is 5.003 - 1.531 = 3.472 g and the ratio NaCl / NaHCO3 = 2.27