Form 1:
Air and Combustion

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To find the composition of active part of air using heated copper turnings.

Procedure

Clamp a completely packed/filled open ended glass tube with copper turnings.

Seal the ends with glass/cotton wool.

Label two graduated syringes as A and B .

Push out air from syringe A.

Pull in air into syringe B.

Attach both syringe A and B on opposite ends of the glass tube.

Determine and record the volume of air in syringe B V1.

Heat the glass tube strongly for about three minutes.

Push all the air slowly from syringe B to syringe A as heating continues. Push all the air slowly from syringe A back to syringe B and repeatedly back and forth.

After about ten minutes, determine the new volume of air in syringe B V2

Set up of apparatus

Sample observations

Colour change from brown to black

Volume of air in syringe B before heating V1 = 158.0cm3

Volume of air in syringe B after heating V2 = 127.2cm3

Volume of air in syringe B used by copper V1 - V2 = 30.8cm3

Sample questions

1. What is the purpose of:

(i) glass/cotton wool

To prevent/stop copper turnings from being blown into the syringe/out of the glass tube

ii) Passing air through the glass tube repeatedly

To ensure all the active part of air is used up

(iii) Passing air through the glass tube slowly

To allow enough time of contact between the active part of and the heated copper turnings

2. State and explain the observations made in the glass tube.

Colour change from brown to black

Brown copper metal reacts with the active part of air/oxygen to form black copper (II) oxide.

Chemical equation

Copper + Oxygen -> Copper (II) oxide
2Cu(s) + O2 (g) -> 2CuO(s)

The reaction reduces the amount/volume of oxygen in syringe “B” leaving the inactive part of air. Copper only react with oxygen when heated.

3. Calculate the % of

(i) Active part of air

% active part of air = V1 - V2/V1 x 100% => 30.8cm3/ 158.0cm3 x 100% = 19.493%

(ii) Inactive part of air

Method 1

% inactive part of air = V2/V1 x 100% =>127.2cm3/158.0cm3 x 100% = 80.506%

Method 2

% inactive part of air = 100% -% active part of air => 100 % - 19.493 % = 80.507%

4. The % of active part of air is theoretically higher than the above while % of inactive part of air is theoretically lower than the above. Explain.

Not all the active part of air reacted with copper

5. State the main gases that constitute:

(a )active part of air. Oxygen

(b) Inactive part of air

Nitrogen, carbon (IV) oxide and noble gases

6. If the copper turnings are replaced with magnesium shavings the % of active part of air obtained is extraordinary very high. Explain.

Magnesium is more reactive than copper. The reaction is highly exothermic. It generates enough heat for magnesium to react with both oxygen and nitrogen in the air.

A white solid/ash mixture of Magnesium oxide and Magnesium nitride is formed.

This considerably reduces the volume of air left after the experiment.

Chemical equation

Magnesium + Oxygen -> magnesium (II) oxide
2Mg(s) + O2 (g) -> 2MgO(s)

Magnesium + Nitrogen -> magnesium (II) nitride
3Mg(s) + N2 (g) -> Mg3N2 (s)

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