Stoichiometry Questions and Answers
Modbury High School
SACE Stage 1 Chemistry
Topic 5 Mole Concept and Stoichiometry
Assignment 5: Volumetric analysis (titrations), stoichiometry SOLUTIONS
Note: Write answers neatly and legibly in your exercise book or on pad paper.
ALWAYS include a title and name for your work and clearly indicate each answer.
1. a) 23.08 and 23.00 mL are concordant titre values. Average titre = (23.08 + 23.00) = 23.04 mL 2
b) Ca(OH)2 + 2HCl → CaCl2 + 2H2O V = 23.04 mL C = 0.0934 M = 0.02304 L V = 20.0 mL C = ? = 0.0200 L unknown known
nHCl = C.V = 0.0934 x 0.0200 = 0.00187 mol nCa(OH)2/nHCl = 1/2 ∴ nCa(OH)2 = 1/2 x nHCl = 1/2 x 0.00187 = 9.35 x 10-4 mol ∴ CCa(OH)2 = n/V = 9.35 x 10-4/0.02304 = 0.406 M (3 sf)
c) Burette
2.
a) CH3COOH + NaOH → NaCH3COO + H2O C = 0.100 M V = 40.25 mL = 0.04025 L V = 50.0 mL = 0.0500 L C = ? known unknown
nCH3COOH = C.V = 0.100 x 0.05000 = 0.005 mol nNaOH/nCH3COOH = 1/1 ∴ nNaOH = nCH3COOH = 0.005 mol ∴ CNaOH = n/V = 0.005/0.04025 = 0.124 M (3 sf)
b) Volumetric pipette
c)
– rinsed with distilled water first to remove any impurities. Ensured all of the inner suface of the pipette was rinsed. – then rinsed with solution to be transferred i.e. acetic acid solution.
– filled up the pipette (with the tip well into the solution) so that the bottom of the meniscus was sitting on the calibration mark.
– transferred solution into conical flask, resting the tip on the inner neck of the flask
SACE Stage 1 Chemistry
Topic 5 Mole Concept and Stoichiometry
Assignment 5: Volumetric analysis (titrations), stoichiometry SOLUTIONS
Note: Write answers neatly and legibly in your exercise book or on pad paper.
ALWAYS include a title and name for your work and clearly indicate each answer.
1. a) 23.08 and 23.00 mL are concordant titre values. Average titre = (23.08 + 23.00) = 23.04 mL 2
b) Ca(OH)2 + 2HCl → CaCl2 + 2H2O V = 23.04 mL C = 0.0934 M = 0.02304 L V = 20.0 mL C = ? = 0.0200 L unknown known
nHCl = C.V = 0.0934 x 0.0200 = 0.00187 mol nCa(OH)2/nHCl = 1/2 ∴ nCa(OH)2 = 1/2 x nHCl = 1/2 x 0.00187 = 9.35 x 10-4 mol ∴ CCa(OH)2 = n/V = 9.35 x 10-4/0.02304 = 0.406 M (3 sf)
c) Burette
2.
a) CH3COOH + NaOH → NaCH3COO + H2O C = 0.100 M V = 40.25 mL = 0.04025 L V = 50.0 mL = 0.0500 L C = ? known unknown
nCH3COOH = C.V = 0.100 x 0.05000 = 0.005 mol nNaOH/nCH3COOH = 1/1 ∴ nNaOH = nCH3COOH = 0.005 mol ∴ CNaOH = n/V = 0.005/0.04025 = 0.124 M (3 sf)
b) Volumetric pipette
c)
– rinsed with distilled water first to remove any impurities. Ensured all of the inner suface of the pipette was rinsed. – then rinsed with solution to be transferred i.e. acetic acid solution.
– filled up the pipette (with the tip well into the solution) so that the bottom of the meniscus was sitting on the calibration mark.
– transferred solution into conical flask, resting the tip on the inner neck of the flask