Lesson: Chemical Reactions and Equations

# Question 1

Why should a magnesium ribbon be cleaned before burning in air?

## Solution:

Magnesium is a highly reactive metal. It reacts with oxygen to form a layer of magnesium oxide on its surface. This layer is quite stable and prevents further reaction of magnesium with oxygen. Hence, the magnesium ribbon is cleaned to remove this layer.

# Question2

Write the balanced equation for the following chemical reactions.

(i)

(ii)

(iii)

## Solution:

(i) ${H}_{2}\left(g\right)+C{l}_{2}\left(g\right)\to 2\text{\hspace{0.17em}}HCl\left(g\right)$

(ii) $3BaC{l}_{2}\left(s\right)+A{l}_{2}{\left(S{O}_{4}\right)}_{3}\left(s\right)\to 3BaS{O}_{4}\left(s\right)+2AlC{l}_{3}\left(s\right)$

(iii) $2\text{\hspace{0.17em}}Na\left(s\right)+2\text{\hspace{0.17em}}{H}_{2}O\left(l\right)\to 2\text{\hspace{0.17em}}NaOH\left(aq\right)+{H}_{2}\left(g\right)$

# Question 3

Write a balanced chemical equation with state symbols for the following reactions.

i.            Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.

ii.            Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

## Solution:

(i) $BaC{l}_{2}\left(aq\right)+N{a}_{2}S{O}_{4}\left(aq\right)\to BaS{O}_{4}\left(s\right)+2\text{\hspace{0.17em}}NaCl\left(aq\right)$

(ii) $NaOH\left(aq\right)+HCl\left(aq\right)\to NaCl\left(aq\right)+{H}_{2}O\left(l\right)$

# Question 4

A solution of a substance ‘X’ is used for white washing.

(i) Name the substance ‘X’ and write its formula.

(ii) Write the reaction of the substance ‘X’ named in (i) above with water.

## Solution:

(i) The substance ‘X’ is calcium oxide. Its chemical formula is CaO.
(ii)
$CaO\left(s\right)+{H}_{2}O\left(l\right)\to Ca{\left(OH\right)}_{2}\left(aq\right)$

# Question 5

Why is the amount of gas collected in one of the test tubes in Activity1.7 (electrolysis of water) double of the amount collected in the other? Name this gas.

## Solution:

The given activity is electrolysis of water. In this activity, hydrogen and oxygen are collected in separate test tubes. A molecule of water ( ${H}_{2}O$ ) contains two hydrogen atoms and one oxygen atom. Therefore, the hydrogen and oxygen produced during electrolysis of water is in the ratio 2:1. Thus, the amount of hydrogen gas collected in one of the test tubes is double the amount of oxygen gas collected in the other.

# Question 6

Why does the colour of copper sulphate solution change when an iron nail is dipped in it?

## Solution:

Iron is more reactive than copper. When an iron nail is dipped in a copper sulphate solution, which is blue in colour, iron displaces copper from the solution and forms ferrous sulphate, which is green in colour.

$Fe\left(s\right)+CuS{O}_{4}\left(aq\right)\to FeS{O}_{4}\left(aq\right)+Cu\left(s\right)$

Therefore, the colour of the copper sulphate solution changes from blue to green.

# Question 7

Give an example of a double displacement reaction.

## Solution:

Sodium carbonate reacts with calcium chloride to form calcium carbonate and sodium chloride. This is a double displacement reaction.

$N{a}_{2}C{O}_{3}\left(aq\right)+CaC{l}_{2}\left(aq\right)\to CaC{O}_{3}\left(s\right)+2NaCl\left(aq\right)$

# Question 8

Identify the substances that are oxidised and the substances that are reduced in the following reactions.

(i) $4\text{\hspace{0.17em}}Na\left(s\right)+{O}_{2}\left(g\right)\to 2\text{\hspace{0.17em}}N{a}_{2}O\left(s\right)$

(ii) $CuO\left(s\right)+{H}_{2}\left(g\right)\to Cu\left(s\right)+{H}_{2}O\left(l\right)$

## Solution:

(i) Sodium (Na) is oxidised as it gains oxygen and oxygen gets reduced.

(ii) Copper oxide (CuO) is reduced to copper (Cu) while hydrogen ( ${H}_{2}$ ) gets oxidised to water ( ${H}_{2}O$ ).