C3 Quantitative Chemistry (Part 2)

Question 1

Formula for concentration

Answer 1

Concentration = mass of solute / volume of solvent

Concentration (g per dm3) = Mass (g)/Volume (dm3)

Concentration (mol per dm3) = Mr of moles/volume (dm3)

Question 2

31.0 cm3 of potassium hydroxide solution neutralised 25.0 cm3 of 2.0 moldm−3 nitric acid.
HNO3 + KOH → KNO3 + H2O
Calculate the concentration of the potassium hydroxide solution in moldm−3

Answer 2

Step 1: Calculate the moles of HNO3
used = Concentration x volume
2 x 0.025 dm3
(25/1000 to convert the units) = 0.05 moles

Step 2 : Calculate the moles of KOH
Ratio is 1:1 therefore number of moles of KOH = 0.05

Step 3 : Calculate the concentration of KOH
Volume = Moles/concentration; 0.05 / 0.031 = 1.61

Question 3

Why is it not always possible to
obtain the theoretical amount of
product in a chemical reaction?

Answer 3

● The reaction may not go to completion because it is reversible.
● Some of the product may be lost when it is separated from the
reaction mixture.
● Some of the reactants may react in ways different to the expected
reaction (side reactions may occur).

Question 4

How is the percentage yield of

a product in a chemical reaction calculated?

Answer 4

%Yield = (Actual mass of a product / Maxium theoretical mass of a product) x 100

Question 5

Calculate the percentage yield from the
following data
Actual yield: 45, 12, 8
Predicted: 100, 50, 40

Answer 5

45, 24, 20

Question 6

What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2?

Answer 6

Step 1 - Write the balanced equation
N2 + 3 H2 → 2 NH3

Step 2 - Calculate the theoretical amount of NH3 .Moles NH3 (ratio of H2 to NH3 is 3:2) of 20/1.5 = 13.3 moles
13.3 X 17 (Mr of NH3) = 227

Step 3 - Calculate percentage yield of NH3
40.5/227 x 100 = 17.8%