Question

Tom was heating a substance $X$ in a china dish which melted into a brown colour substance and he placed a glass sheet on it which became foggy after a while. On further heating he saw a black powdery substance $Y$ left on the china dish.
-Name $X$ and $Y$
-Write the chemical reaction of this process.

Answer 1

Hint : Heating generally causes oxidation reaction to occur. We know that air is composed of $78\%$ nitrogen and $21\%$ oxygen and the rest $1\%$ is other gases. Metal usually combines with oxygen to form metal oxide as they don’t combine with nitrogen. So, when metal combines with oxygen the reaction is then known as oxidation reaction.

Complete Step By Step Answer:
The metal which is combining here to form a metal oxide is the copper metal.
Copper combines with oxygen to form copper oxide
the oxidation reaction will be:
$2Cu(s) + {O_2}(g) \to 2CuO(s)$
When copper is subjected to heating in a china dish, the surface of the copper seems to be coated with a black coloured substance which is possibly due to the occurrence of surface oxidation. Copper metal melts gradually upon heating and turns into a brown colour substance. On further heating, a black powdery substance i.e. copper (II) oxide is formed upon surface oxidation. Actually, copper has one free electron per atom so when copper is exposed to heat the oxygen molecules present in the air combines with the free electrons of copper and hence forms the molecule of copper oxide. If only one atom of carbon combines with oxygen, then it is known as cupric oxide and whereas if two atoms of carbon will combine with oxygen then it will be known as cuprous oxide. Hence, we see that the oxide formed here is the cupric oxide.
Chemical reaction of the process is known as redox reaction which is shown below: $2Cu(s) + {O_2}(g) \to 2CuO(s)$
In this reaction, oxygen is being reduced while copper is being oxidized.
So, the answer to the above question is
$X =$ copper
$Y =$ cupric oxide or copper $(II)$ oxide.

Note :
The black coloured cupric oxide formed can be reduced to form copper metal again.
The reduction reaction is: $CuO(s) + {H_2}(g) \to Cu(s) + {H_2}O(l)$
Cupric oxide is considered to be fully oxidized while cuprous oxide is still in an active state.