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The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 2
QUESTION 1 A sample of an ore was analysed for Cu2+ as follows. A 1.25 gram sample of the ore was dissolved in acid and diluted to volume in a 250 mL volumetric flask. A 20 mL portion of the resulting solution was transferred by pipette to a 50 mL volumetric flask and diluted to volume. An analysis of this solution gave the concentration of Cu2+ as 4.62 µg/L. What is the weight percent of Cu in the original ore? Solution
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 3
QUESTION 2 The amount of oxalic acid (H2C2O4) in a sample of rhubarb was determined by reacting with Fe3+. After extracting a 10.62 g of rhubarb with a solvent, oxidation of the oxalic acid required 36.44 mL of 0.0130 M Fe3+. What is the weight percent of oxalic acid in the sample of rhubarb?
3 2
( ) 2 2 4( ) 2 ( ) ( ) 2( ) 3 ( )2 2 2 2 2aq aq l aq g aqFe H C O H O Fe CO H O
Solution
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 4
QUESTION 3 An analysis for disulfiram, C10H20N2S4, in Antabuse is carried out by oxidising the sulfur to H2SO4 and titrating the H2SO4 with NaOH. If a 0.4613 g sample of Antabuse is taken through this process and requires 34.85 mL of 0.02500 M NaOH to titrate the H2SO4, what is the % (w/w) disulfiram in the sample? Solution
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 5
QUESTION 4 A chemist wishes to analyse the salt (sodium chloride) content of two brands of “light” potato chips. He enlists the help of a group of students in Year 12. The students decide to crush the
250.00g bag of each brand of chips up and they then dissolve a 25.00 g sample of each
brand in separate 100.00ml volumetric flasks. Three 20.00ml aliquots are removed from
each of the volumetric flasks and placed into conical flasks that contain an excess amount of silver nitrate. A precipitate of silver chloride is formed. The precipitate formed in each conical flask is filtered, washed and dried until a constant mass is obtained. The average mass of precipitate for Brand A was 0.123g and the average mass for Brand B was 0.211g .
(a) What was the average salt content in the 20.00ml samples of Brand A and Brand B?
4 marks
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 6
(b) What was the average salt content, in grams, in each 250.00g packet of chips?
2 marks (c) What should be quoted on the back of each chip packet regarding the salt content per 100 g sample? State your answer for Brand A in terms of %w/w and Brand B in
terms of parts per million.
2 marks (d) What errors may result in the calculated salt content being higher than that specified by the manufacturer? Give two examples.
2 marks
Total 10 marks
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 7
QUESTION 5
Sodium hydrogen sulfate, 4NaHSO , is an acidic solid that is often used to adjust the pH of
swimming pools. The purity of one brand of this material was determined by titration with a
solution of sodium carbonate, 2 3Na CO by the following method.
An aqueous solution of the pool acid was prepared by dissolving 2.15 g of the material in
water and making up the volume to 250 mL in a volumetric flask.
1.082 g of anhydrous 2 3Na CO was dissolved in water and made up to 250 mL in a second
volumetric flask. After thorough mixing, 20.00 mL aliquots of this solution were pipetted into
a flask, and titrated with the solution of pool acid. An average titre of 23.42 mL was needed
to neutralise the base. The equation for the reaction is
4( ) 2 3( ) 2 4( ) 2( ) 2 ( )2 2aq aq aq g lNaHSO Na CO Na SO CO H O
(a) How many mole of acid was in the average titre?
3 marks
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 8
(b) Calculate the number of mole of 4NaHSO that were in the first 250 mL flask.
1 mark (c) Calculate the percentage purity of the pool acid.
2 marks
Total 6 marks
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 9
QUESTION 6 A sample of commercial vinegar is to be analysed for the concentration of ethanoic acid,
3CH COOH . The student firstly transferred 20.00ml of the vinegar to a volumetric flask and
diluted it to a volume of 100ml . A 20.00ml sample of this diluted vinegar solution was then
transferred to a conical flask, and titrated against a 0.7432M solution of sodium hydroxide.
This process was repeated four times, and the titres obtained are show in the table below.
Titre No. Volume ( ml )
1
20.45 2 19.35 3 18.40 4 18.65 5
18.60
(a) Write an equation for the reaction between ethanoic acid and sodium hydroxide.
1 mark
(b) Which titres should be used in your calculations? Give a reason for your answer.
2 marks
(c) (i) Calculate the concentration of the original vinegar solution.
3 marks
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 10
(ii) State this concentration in terms of 1gml and hence determine the mass of
ethanoic acid in 20.00ml of commercial vinegar.
3 marks
(d) If the volumetric flask was washed with water prior to its use and not dried, would the calculated concentration of vinegar be higher, lower or the same as your calculated value?
1 mark
Total 10 marks
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 11
QUESTION 7 – VCAA 2013 SAMPLE EXAM QUESTION Phosphorus is an essential ingredient in plant fertiliser. The phosphorus content in fertiliser
can be determined as a percentage, by mass, of 2 5PO . A 3.256 g sample of fertiliser is mixed
with 40.0 mL of deionised water and the insoluble residue is then removed using vacuum
filtration. 45.0 mL of 10% 4 2.7MgSO H O solution is added to the filtrate followed by 150.0 mL
of 2 M 3NH solution. A white precipitate forms. This precipitate is filtered and washed with
three 5 mL portions of deionised water. The final mass of the precipitate, once thoroughly
dried, was 4.141 g. The formula of the precipitate is known to be 4 4 2.6MgNH PO H O . Assume
that the experiment was conducted at 25 °C and that all the phosphorus had been precipitated
as 4 4 2.6MgNH PO H O . Calculate the mass of 2 5PO in 1.00 kg of fertiliser.
(Molar mass of 4 4 2.6MgNH PO H O = 245.3 g mol–1. Molar mass of 2 5PO = 142.0 g mol–1.)
Solution
4 marks
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 12
QUESTION 8
100 L of air at SLC is slowly bubbled through 200 mL of 0.0300M 2)(OHBa solution,
which is present in excess. 2)(OHBa reacts with 2CO in air according to the following
equation:
)(2)(3)(2)(2)( lsgaq OHBaCOCOOHBa
The precipitate is filtered and removed. A few drops of phenolphthalein are added to the
filtrate turning it pink. If 25.00 mL of 0.200M HCl is needed to neutralise the solution,
calculate the percentage by volume of 2CO in the air.
Solution
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 13
QUESTION 1
QUESTION 2
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 14
QUESTION 3
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 15
QUESTION 4
(a)
(b)
(c)
Answer is 1.00% /w w
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 16
(d) Co-precipitation of other ions. Incomplete washing. DO NOT state incomplete drying – this would not be accepted in the exams. If the question states that a precipitate is dried “to a constant mass”, you must assume that the drying process was completed correctly, and that no water remains.
QUESTION 5
4( ) 2 3( ) 2 4( ) 2( ) 2 ( )2 2aq aq aq g lNaHSO Na CO Na SO CO H O
(a)
(b)
(c)
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 17
QUESTION 6
(a) )1(233 )()()( OHaqCOONaCHaqCOOHCHaqNaOH
(b) Use titres 4 and 5.
We only use values that are concordant (within 0.10 mL of one another) since the other values may have arisen from under/overshooting the endpoint.
(c) (i) Average titre 18.65 18.60
18.6252
ml
18.63 ml
18.625
( ) 0.7432 0.01384211000
n NaOH cV mol
3( ) ( ) 0.0138421n NaOH n CH COOH mol
(This is the amount, in mole, in 20 ml of diluted vinegar)
In 100 ml of diluted vinegar: 100
0.0138421 0.069210520
mol mol
This is also the mol in the original 20.00 ml of undiluted vinegar
3
0.06921053.460525 3.461
0.020
nCH COOH M M
V
(ii) Always use the unrounded values in subsequent calculations.
3.460525 3.460525 / 3.460525 /1000 0.003460525 /1M mol L mol ml mol ml
0.003460525 60 0.2076315m n Mr g
i.e. 0.2076315 /g ml
i.e. 0.2076315 20 4.15263 / 20g ml
Therefore, in 20 mL there are 4.153 g ethanoic acid.
(d) Unchanged as any additional water would not change the number of mole of ethanoic acid, and hence the titres obtained would not be affected.
The School For Excellence 2017 Unit 3 & 4 Chemistry – Stoichiometry – Worksheet 1 Page 18
QUESTION 7