Question

Plastic cup C

Mass of cup, water, and stirrer: 57.24g

Mass of sodium bicarbonate: 2.02g

Mass of citric acid: .76g

Total mass: 60.02g

Mass of cup, the solution, and stirrer after reaction: 59.53g

Difference: 0.49g

H₃C₆H₅O₇(aq) + 3NaHCO₃(aq) --->Na₃C₆H₅O₇(aq) + 3H₂O(l) + 3CO₂(g)

1)Determine which reactant is the limiting reactant in the plastic up C. Describe your reasoning

2) Calculate the theoretical yield of carbon dioxide in plastic cup C

3) Calculate the percentage yield in plastic cup C

Answer 1

Given:

Mass of citric acid = 0.76 g

Mass of sodium bicarbonate = 2.02 g

Mass difference = 0.49 g= mass of CO2 (g)

1). Limiting reactant determination:

Calculation of moles of NaHCO3 = Mass of NaHCO3 / Molar mass of NaHCO3

= 2.02 g / 84.007 g per mol

= 0.02404 mol

Calculation of moles of citric acid

Moles of citric acid = 0.76 g / 192.124 g per mol

= 0.00396 mol citric acid

Calculation of moles of product (CO2 ) from both reactant :

a). From citric acid ;

mol ratio of CO2 : Citric acid is 1 : 3

moles of CO2 = Moles of Citric acid x 3 moles of CO3 / 1 mol citric acid

= 0.00396 mol x 3 mol CO2 / 1 mol citric acid

= 0.01187 mol CO2

b). From moles of NaHCO3

moles of CO2 = 0.02402 mol NaHCO3 x 3 mol CO3 / 3 mol NaHCO3

= 0.02402 mol CO2

Moles of CO2 from moles of citric acid are less than from Sodium bicarbonate so

Citric acid is the limiting reactant

2).

Calculation of theoretical yield :

Mass of product calculated by using moles of product obtained from limiting reactant

Mass of CO2 = 0.01187 mol x molar mass of CO2

= 0.01187 mol x 44.009 g /mol

= 0.5223 g

Theoretical yield = 0.5223 g

3).

Percent yield

= (Actual mass / Theoretical mass ) x 100

= (0.49 / 0.5223 g )x 100

= 93.8 %

Percent yield = 93.8 %