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Stoichiometry Questions and Answers (Page 2)

Stoichiometry Questions and Answers(Page 2)

Question 1

Sodium sulphate has the following composition by mass: sodium 32.4%, sulphur 22.5% and oxygen 45.1%. What is its empirical formula?

Solution 1

SodiumSulphurOxygen
Mass32.422.545.1
Moles32.4/2322.5/3245.1/16
= 1.41= 0.7= 2.82
Molar Ratio1.41/0.70.7/0.72.82/0.7
= 2= 1= 4

Therefore the empirical formulae = Na2SO4



Question 2

a. Calculate the empirical formula of an organic liquid containing 26.67% of carbon and 2.22% of hydrogen, with the rest being oxygen.
b. The Mr of the liquid is 90. What is its molecular formula?

Solution 2

a. first we find the percentage by mass of oxygen
% Oxygen = 100 – 26.67 – 2.22
= 71.11%

CarbonHydrogenOxygen
Mass26.672.2271.11
Moles26.67/122.22/171.11/16
= 2.22= 2.22= 4.44
Molar Ratio2.22/2.222.22/2.224.44/2.22
= 1= 1= 2

Therefore the empirical formula = CHO2

b. To find molecular formula we multiply empirical formula by number of formula units.

Formula units = Molecular mass ÷ Empirical mass
= 90 ÷ (12 + 1 + 32)
= 90 ÷ 45
= 2

Molecular formula = 2(CHO2)
= C2H2O4



Question 3

Iron is extracted from its ore, haematite, in the blast furnace. The main extraction reaction is:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2
(g)
a. Name the reducing agent in this process.
b. Name the oxide of iron shown in the equation.
c. Explain why this is a redox reaction.
d. Calculate the mass of iron which will be produced from 640 tonnes of haematite.

Solution 3

a. as you can see from the equation, the oxygen lost by Fe2O3 was gained by 3CO.
CO is the reducing agent

b. Iron (iii) Oxide

c. This is a redox equation because Fe2O3 is reduced whilst CO is oxidised.

d. When facing questions like this, the first thing is to find the molar ratios of reactants and products. On a chemical equation, the molar ratio is the ratio of the coefficients of the reactants and the products.

Fe2O3(s) : 3CO(g) : 2Fe : 3CO2
(g)
= 1:3:2:3
In short, all that it means is, 1 mole of iron(iii) oxide reacts with 3 moles of carbon monoxide to produce 2 moles of iron and 3 moles of carbon dioxide.

Now lets answer the question

  • convert 640 tonnes of haematite to grams
  • = 640 × 1 000 000
  • = 640 000 000g

Next find the number of moles of haematite

  • Number of moles = mass ÷ molar mass
  • = 640 000 000/(56 × 2 + 16 × 3)
  • = 640 000 000/(160)
  • = 4 000 000 moles

Now lets use the molar ratio to find the moles of iron produced:

From the equation Fe2O3(s) : Fe
= 1:2
= 4 000 000:x

x/2 = 4 000 000/1
x = 4 000 000(2)
x = 8 000 000 moles of Fe produced.

Finally lets convert the number of moles of Fe to mass:

Mass = number of moles × Mr
= 8 000 000 × 56
= 448 000 000 g
= 448 tonnes



Question 4

Consider the following information about the newly discovered element, vulcium, whose symbol is Vu.

"Vulcium is a solid at room temperature. It is easily cut
by a penknife to reveal a shiny surface which tarnishes
quite rapidly. It reacts violently with water, liberating a
flammable gas and forms a solution with a pH of 13.
When vulcium reacts with chlorine, it forms a white crystalline solid containing 29.5% chlorine."

(Ar : Vu = 85)

a. Calculate the empirical formula of vulcium chloride.
b. To which group of the Periodic Table should vulcium be assigned?
c. Write a word and balanced chemical equation for the reaction between vulcium and chlorine.
d. What other information in the description supports the assignment of group you have given to vulcium?
e. What type of bonding is present in vulcium chloride?
f. Write a word and balanced chemical equation for the reaction between vulcium and water.

Solution 4

a. % of Vulcium = 100 – 29.5
= 70.5%

VulciumChlorine
Mass70.529.5
Molar mass8535.5
Number of moles70.5/85= 0.8329.5/35.5 = 0.83
Molar ratio11

Empirical formula = VuCl

b. Group 1
because the fact that Vu bonds with Cl which has a dificiency of 1 electorn in the outershell in the ratio 1:1 means that Vu has 1 electron in the outershell. So its Group 1 because group number is equal to the number of electrons in the outershell.

c. Vulcium + Chlorine → Vulcium Chloride;
2Vu + Cl2 → 2VuCl

d. Like all other Group 1 elements:

  • It is easily cut by a penknife to reveal a shiny surface.
  • It tarnishes quite rapidly.
  • It reacts violently with water, liberating a flammable gas and forms a solution with a pH of 13.

e. Ionic bond
Group 1 elements form ionic bonds with halogens.

f. Vulcium + water → Vulcium hydroxide + hydrogen
2Vu + 2H2O → 2VuOH + H2



Question 5

0.048 g of magnesium was reacted with excess dilute hydrochloric acid at room temperature and pressure. The hydrogen gas given off was collected.
a. Write a word and balanced symbol equation for the reaction taking place.
b. Draw a diagram of an apparatus which could be used to carry out this experiment and collect the hydrogen gas.
c. How many moles of magnesium were used?
d. Using the equation you have written in your answer to a, calculate the number of moles of hydrogen and hence the volume of this gas produced.

Solution 5

a. Magnesium + Hydrochloric acid → Magnesium Chloride + Hydrogen
Mg + 2HCl → MgCl2 + H2

b.
Hcl+Mg experiment

c. Number of moles = mass/molar mass
= 0.048/24
= 0.002 moles

d. Molar ratio Mg: H = 1:2
= 0.002:x
x = 0.004 moles of Hydrogen

Volume = number of moles × molar volume
= 0.004 × 28
= 0.112 dm3



Question 6

Magnesium burns in chlorine to give magnesium chloride, MgCl2. In an experiment, 24 g of magnesium was found to react with 71 g of chlorine.
a. How much magnesium chloride was obtained in the experiment?
b. How much chlorine will react with 12 g of magnesium?
c. How much magnesium chloride will form, in b?

Solution 6

To be Continued



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