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Sol Inn q&a

In a neutralisation experiment, 25 cm3 of dilute sulfuric acid was required to react completely
with 40 cm3 of a solution of 0.25 mol dm−3 potassium hydroxide.
a. Write a balanced chemical equation for the reaction between dilute sulfuric acid and
potassium hydroxide.
H2SO4 (aq) + 2 KOH (aq) --------οƒ  K2SO4 (aq) + 2 H2O (l)
b. Calculate the number of moles of potassium hydroxide solution used in the reaction.
π‘šπ‘œπ‘™π‘’π‘  (𝑛)
Formula: π‘€π‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ (𝑀) = π‘‰π‘œπ‘™π‘’π‘šπ‘’ (𝑉)
From above molarity equation, we get
π‘šπ‘œπ‘™π‘’π‘  = π‘€π‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ ∗ π‘‰π‘œπ‘™π‘’π‘šπ‘’
𝑛 =𝑀∗𝑉
Molarity of potassium hydroxide (M) = 0.25 mol/dm3
Volume of potassium hydroxide (V) = 40 cm3 = 0.040 dm3, (1cm3 = 0.001 dm3)
Moles of potassium hydroxide =??
𝑛 = 0.25
∗ 0.040 π‘‘π‘š3
𝑛 = 0.01 π‘šπ‘œπ‘™π‘ 
c. How many moles of dilute sulfuric acid would this number of moles of potassium
hydroxide react with?
From the balanced chemical equation the mole ratio of sulphuric acid (H2SO4) to
potassium hydroxide (KOH) is 1:2
0.01 π‘šπ‘œπ‘™π‘  𝐾𝑂𝐻 ∗
1 π‘šπ‘œπ‘™ 𝐻2𝑆𝑂4
2 π‘šπ‘œπ‘™π‘  𝐾𝑂𝐻
= 0.005 π‘šπ‘œπ‘™π‘  𝐻2𝑆𝑂4
d. Calculate the concentration of the dilute sulphuric acid.
Moles of sulfuric acid (n) = 0.005 mols
Volume of sulphuric acid (V) = 25 cm3 = 0.025 dm3
0.005 π‘šπ‘œπ‘™π‘ 
= 0.2 𝑀
0.025 π‘‘π‘š3
e. Which indicator could have been used to determine when neutralisation had just
Phenolphthalein indicator is used
A type of coal contains 0.5% of sulfur by mass.
a. Write an equation for the formation of sulfur dioxide gas when this coal is burned.
S (in coal) + O2 (g) -------οƒ  SO2 (g)
32.0 g
32.0 g
64.0 g
b. What mass of sulfur is contained in 1500 tonnes of coal?
1500 π‘‘π‘œπ‘›π‘›π‘’π‘  ∗
= 7.5 π‘‘π‘œπ‘›π‘›π‘’π‘  π‘œπ‘“ π‘ π‘’π‘™π‘“π‘’π‘Ÿ
b. What mass of sulfur dioxide gas would be formed if 1500 tonnes of coal were burned?
Since we have 7.5 tonnes of sulphur in 1500 tonnes of coal, when the 7.5 tonnes of
sulphur is burned
Mass of SO2 formed:
7.5 π‘‘π‘œπ‘›π‘›π‘’π‘  π‘†π‘’π‘™π‘“π‘’π‘Ÿ ∗
106 𝑔
64 𝑔 𝑆𝑂2
= 15 ∗ 106 𝑔 π‘œπ‘“ 𝑆𝑂2
1 π‘‘π‘œπ‘›π‘›π‘’ 32 𝑔 π‘ π‘’π‘™π‘“π‘’π‘Ÿ
c. What volume would this mass of sulfur dioxide gas occupy, measured at room
temperature and pressure (rtp)? (Ar: O = 16; S = 32. One mole of a gas occupies 24
dm3 at rtp.)
15 ∗ 106 𝑔 π‘œπ‘“ 𝑆𝑂2 ∗
1 π‘šπ‘œπ‘™π‘’ 𝑆𝑂2
24 π‘‘π‘š3
= 5.6 ∗ 106 π‘‘π‘š3
64 𝑔 𝑆𝑂2
1 π‘šπ‘œπ‘™π‘’ 𝑆𝑂2
a. Copy and complete the above table by calculating the percentage of nitrogen in the fertilisers
sodium nitrate and potassium nitrate.
(Ar: H = 1; N = 14; O = 16; Na = 23; K = 39; Ca = 40)
Percentage of nitrogen in Sodium nitrate (NaNO3):
Molarmass of NaNO3 = 23 + 14 + (3*16) = 23 + 14 + 48 = 85 g/mol
Mass of Nitrogen in 1 mole of NaNO3 = 14 g
% π‘œπ‘“ 𝑁 𝑖𝑛 π‘π‘Žπ‘π‘‚3 =
∗ 100 = 16.47 %
Percentage of nitrogen in Potassium nitrate (KNO3):
Molarmass of KNO3 = 39 + 14 + (3*16) = 39 + 14 + 48 = 101 g/mol
Mass of Nitrogen in 1 mole of KNO3 = 39 g
% π‘œπ‘“ 𝑁 𝑖𝑛 𝐾𝑁𝑂3 =
∗ 100 = 38.61 %
b. Including the data you have just calculated, which of the fertilisers contains:
(i) The largest percentage of nitrogen?
Ammonia solution (NH3) with 82.4% Nitrogen
(ii) The smallest percentage of nitrogen?
Sodium nitrate (NaNO3) with 16.5% Nitrogen
d. Give the chemical name for the fertiliser that goes by the name Nitram®.
Ammonium nitrate (NH4NO3)
e. Ammonia can be used directly as a fertiliser but not very commonly. Think of two
reasons why ammonia is not often used directly as a fertiliser.
i). Very high nitrogen percentage in ammonia
ii). Aquatic life is harmed by ammonia even at extreme low concentrations.
e. Nitram® fertiliser is manufactured by the reaction of nitric acid with ammonia solution
according to the equation:
NH3(aq) + HNO3(aq) → NH4NO3(aq)
A bag of Nitram® may contain 50 kg of ammonium nitrate. What mass of nitric acid would be
required to make it?
103 𝑔
1 π‘šπ‘œπ‘™π‘’ 𝑁𝐻4𝑁𝑂3
1 π‘šπ‘œπ‘™π‘’ 𝐻𝑁𝑂3
63 𝑔 𝐻𝑁𝑂3
50 𝐾𝑔 𝑁𝐻4𝑁𝑂3 ∗
1 𝐾𝑔
80 𝑔 𝑁𝐻4𝑁𝑂3
1 π‘šπ‘œπ‘™π‘’ 𝑁𝐻4𝑁𝑂3 1 π‘šπ‘œπ‘™π‘’ 𝐻𝑁𝑂3
= 39375 𝑔 = 39.4 𝐾𝑔
The following diagram shows the three states of matter and how they can be interchanged.
a. Name the changes A to E.
A = Freezing
B = Melting
C = Condensation
D = Vaporization
E = Sublimation
b. Name a substance which will undergo change E.
Iodine, Camphor …etc
c. Name a substance which will undergo changes from solid to liquid to gas between 0 °C
and 100 °C.
d. Describe what happens to the particles of the solid during change E.
During the sublimation process, the particles of the solid absorb energy to move fast
and overcome the inter-particle attractions to become gas.
e. Which of the changes A to E will involve:
(i) An input of heat energy?
B, D, E
(ii) An output of heat energy?
A, C, E
The table below shows the melting points, boiling points and densities of substances A to D.
a. Which substance is a gas at room temperature?
b. Which substance is a liquid at room temperature?
c. Which substances are solids at room temperature?
A and C
d. Which substance is most likely to be a metal?
e. Which substance will be a liquid at −260 °C?
f. What is the melting point of the least dense nonmetal?
-266 oC
g. Which substances are gases at 72 °C?
B and D
Bby questions: