Dry Carbon (II) Oxide gas reacts with heated Lead (II) as shown in the equation below. PbO(s) + CO(g) → CO2(g) + Pb(s) a) Name the process undergone by the Lead (II) Oxide. b) Give a reason for your answer (a) above. c) Name another gas that can be used to perform the same function as Carbon (II) Oxide gas in the above reaction. 2. The diagram below shows a Bunsen burner when in use. d) Name the region labelled C and D.

In the reaction $\text{PbO(s)} + \text{CO(g)} \rightarrow \text{CO}_2\text{(g)} + \text{Pb(s)}$, Lead (II) Oxide ($\text{PbO}$) loses oxygen to form Lead ($\text{Pb}$).

The loss of oxygen is called reduction.
The oxidation state of Lead changes from +2 in $\text{PbO}$ to 0 in $\text{Pb}$, which is a decrease in oxidation state, indicating reduction.

The process undergone by the Lead (II) Oxide is reduction.

b)
Reason: Lead (II) Oxide loses oxygen atoms to form Lead metal.

c)
Another gas that can be used to perform the same function as Carbon (II) Oxide (a reducing agent) is hydrogen gas ($\text{H}_2$).

2. d)
Name the region labelled C and D.
C: Non-luminous zone (or Outer zone)
D: Luminous zone (or Inner blue cone)