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# you are given 1.0MHCl 65.1 liters of it + an unknown mixture of NaCHO3 and NaCl= CO2 which is .802g+H2O+NaCl how do you solve for the ratio of NaCl and NaCHO3

you are given 1.0MHCl 65.1 liters of it + an unknown mixture of NaCHO3 and NaCl= CO2 which is .802g+H2O+NaCl how do you solve for the ratio of NaCl and NaCHO3 in the unknow mixture the uknown mixture weighs 5.003g

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
.802g CO2 *(1mol CO2/ 44.01g)* ( 1 mol C /1 mol CO2) = 0.01822 mol C . since the only Carbon source is bicarbonate, and the Acid is in excess, mol C = mol NaHCO3

.01822 mol NaHCO3 *(84.007g NaHCO3/ 1mol NaHCO3) = 1.5309 g NaHCO3.
therefore ratio is 1.5309/ 5.003 = 0.306 or

30.6% NaHCO3: 69.4% NaCl

Hey Phill -- looks like C in 0.8g CO2 comes from C in NaCHO3 ... C is 12/44 of CO2 = 0.22g NaCHO3: C is 0.22g ... O3 is 0.88g ... NaH is 0.44g -> total is 1.54g ... Since Na mix is 5g, NaCHO3 is 30% and NaCl is the other 70% -> 7:3 ratio ... Best wishes :)