Assuming that sodium bicarbonate is the limiting reactant, determine the percent yield of a reaction between 2.01 g of sodium bicarbonate and 24.6 mL of 1.5 M acetic acid? They produce 0.54 L of 20 degree C CO2 at 750 mmhg.

1 Answer
May 19, 2015

The percent yield for your reaction will be equal to 92.9%.

Start with the balanced chemical equation for the reaction that takes place between sodium bicarbonate, #NaHCO_3#, and acetic acid, #CH_3COOH#.

#NaHCO_(3(s)) + CH_3COOH_((aq)) -> CH_3COONa_((aq)) + H_2O_((l)) + CO_(2(g))#

Notice that you have #1:1# mole ratios between all the species that take part in the reaction. This means that 1 mole of sodium bicarbonate will need 1 mole of acetic acid for the reaction to take place, and will produce 1 mole of carbon dioxide.

Use sodium bicarbonate's molar mass to determine how many moles you have

#2.01cancel("g") * "1 mole"/(84.01cancel("g")) = "0.0239 moles"# #NaHCO_3#

Since you know that sodium bicarbonate acts as a limiting reagent, you can predict that the number of moles of acetic acid will be bigger than 0.0239.

Use the acetic acid solution's molarity to determine how many moles you have

#C = n/V => n = C * V#

#n_(HAc) = "1.5 M" * 24.6 * 10^(-3)"L" = "0.0369 moles"# #HAc#

SIDE NOTE HAc is just another notation used for acetic acid.

Indeed, sodium bciarbonate will act as a limiting reagent, determining the number of moles of acetic acid that react. The theoretic yield for the reaction, i.e. what you would produce if the reaction had a 100% yield, implies that 0.0239 moles of carbon dioxide will be produced.

Use the ideal gas law equation to see how many moles of #CO_2# were actually produced

#PV = nRT => n = (PV)/(RT)#

#n_(CO_2) = (750/760cancel("atm") * 0.54cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 20)cancel("K")) = "0.0222 moles"# #CO_2#

The actual yield of the reaction is 0.0222 moles, which implies that the percent yield will be

#"%yield" = "actual yield"/"theoretical yield" * 100#

#"%yield" = (0.0222cancel("moles"))/(0.0239cancel("moles")) * 100 = color(green)("92.9%")#