Question

9. 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)

a) 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?

b) If 10 L of 0.5 M sodium hydrogen carbonate solution were actually reacted, what is the percent yield of the reaction?

Answer 1

a)

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Tmperature

R = ideal gas constant

V = 1.481*10^3 dL = 148.1 Liters

T = 146.4 F = 63.555556°C

T = 63.555556+273.15 = 336.70 K

P = 681 torr = 681/760 atm = 0.89605 atm

substitute

n = PV/(RT) = (0.89605)(148.1)/(0.082*336.70)

n = 4.806 moles of CO2 obtained

then, relate moles of CO2 to NaHCO3

ratio is 1 mol of CO2 = 1 mol of NaHCO3

4.806 mol of CO2 = 4.806 mol of NaHCO3

change to mass

mass of NaHCO3 = mol*MW = (4.806)(84.007 ) = 403.73 g of NaHCO3

b)

% yield = real / theoretical * 100%

real amount= 4.806 mol of NaHCO3

theoreitcal :

from the Molarity data:

mol = Molarity * Voloume = 10 L * 0.5 mol / L = 5 mol of N

Solve

% yield = 4.806 /5*100 = 96.12%