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.

Question

Wyatt Adkins · Accepted Answer

Your reaction will have a percent yield of 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.

Determine the number of moles you have by using the molar mass of sodium bicarbonate.

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

Acetic acid will have a mole count greater than 0.0239 since you are aware that sodium bicarbonate serves as a limiting reagent.

Calculate the number of moles by using the molarity of the acetic acid solution.

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

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

SIDE NOTE: Another notation for acetic acid is HAc.

The amount of acetic acid that reacts will be limited by the presence of sodium bicarbonate, which also acts as a limiting reagent. Based on the theoretical yield of the reaction—that is, the amount of carbon dioxide that would be produced in the event that the reaction had a 100% yield—0.02339 moles of carbon dioxide are expected to 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#

Since the reaction's actual yield is 0.0222 moles, the yield percentage will be

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

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

Cameron Andrews · Answer

undefined To determine the percent yield, first calculate the theoretical yield of CO2 using stoichiometry. Then, convert the volume of CO2 to moles using the ideal gas law. Finally, use the actual yield provided to calculate the percent yield using the formula:

Percent yield = (actual yield / theoretical yield) * 100%.

The molar mass of sodium bicarbonate (NaHCO3) is 84.01 g/mol. The balanced equation for the reaction is: 
NaHCO3 + CH3COOH -> CO2 + H2O + CH3COONa

Based on the stoichiometry, 1 mole of NaHCO3 produces 1 mole of CO2.

Theoretical yield of CO2 = (mass of NaHCO3 / molar mass of NaHCO3) * (moles of CO2 produced per mole of NaHCO3)

Actual yield is given as 0.54 L of CO2 at 750 mmHg and 20°C. Convert this to moles of CO2 using the ideal gas law:

PV = nRT 
Where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Convert pressure to atm (750 mmHg = 0.987 atm), volume to liters (0.54 L), and temperature to Kelvin (20°C = 293 K). Then solve for moles of CO2.

Once you have the theoretical and actual yields, calculate the percent yield.