Question

Carbonic acid decomposes in warm aqueous solution to produce H2O(l) and CO2(g). Supposing the reaction below:

HCl(aq) + NaHCO3(aq)  NaCl(aq) + CO2(g) + H2O(l)

If this reaction is performed, yielding 1.481x103 dL of carbon dioxide at 146.4˚F and 681 torr, what mass of sodium hydrogen carbonate was initially reacted with the hydrochloric acid? If 10 L of 0.5 M sodium hydrogen carbonate solution were actually reacted, what is the percent yield of the reaction?

Please explain in detail.

Answer 1

1dL=0.1L, 1.481*10³ dL= 1.481*1000*0.1= 1.481*100 L= 148.11L

hence V= 14.81L, given T= 146.4 deg.F, this need to be first converted to deg.C, F=1.8C+32

C= (F-32)/1.8= (146.4-32)/1.8 =63.56 deg.c, this need to be converted to K T= 63.56+273 =336.56K

T= 336.56K, V= 148.1L, P= 681 Torr, 760 torr= 1atm, P= 681/760 atm=0.90 atm, R =0.0821 L.atm/mole.K

mole of CO2 generated ,n= PV/RT= 0.9*148.1/(0.0821*336.56)=4.8 moles

from the reaction NaHCO3(aq)+ HCl(aq) ----------->NaCl(aq)+ CO2(g)+ H2O(l)

1 mole of CO2 is produced from 1 mole of NaHCO3

0.48 moles of CO2 is produced from 0.48 moles of NaHCO3.

molar mass of NaHCO3= 23+1+12+3*16=84 g/mole

mass of NaHCO3 required= moles* molar mass =4.88*84 gm =403.2 gm

2. moles of NaHCO3 in 10L of 0.5M= molarity* volume inL=0.5*10= 5moles

moles of CO2 should have been formed= 5 moles

but moles of CO2 formed= 4.8 moles

% yield= 100* moles of CO2 produced/ moles of CO2 should have been produced = 100*4.8/5 = 96%