Gases: Law of Combining Volumes (Worksheet)

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Page ID: 20175

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Q1

Since 2 vol of acetylene require 5 volumes of oxygen, 500 c.ft will require 500 X 5/2 = 1250 c.ft of oxygen.

For 21 volumes of oxygen we require 100 volumes of air.

∴ for 1250 c.ft of oxygen,

1250 X 100 / 21 = 5952.38 c.ft of air is required

Q4

In a combustion chamber containing 5 L of carbon monoxide and 2.5 L of oxygen is ignited at 298 K and 1 atmosphere pressure. Assuming complete combustion and no loss of gas, what will be the volume of carbon dioxide formed at 298 K and 1 atmosphere pressure?

Solution:

Carbon monoxide combines with oxygen according to the equation CO + O₂ → CO₂.

The balanced equation will be 2CO + O₂ → 2CO₂.

Thus 2 volumes of carbon monoxide and 1 vol of oxygen combine to form 2 volumes of carbon di oxide.

The simple ratio of 2:1:2 which will follow the Gay lussac's law of combining volumes.

since all the gases are at the same temperature and pressure 5 L of carbon monoxide and 2.5 L of oxygen will give 5L of Carbon dioxide.

Q5

Q6

What volume of propane $C_3H_{6(g)}$ reacts with $100 \;cm^3$ of oxygen used in the complete combustion reaction, assuming no changing pressure and temperature: (hint: construct the balanced combustion reaction).

\[ C_3H_{6(g)} + 5O+2 \rightarrow 3CO_2 + 4H-2O \nonumber \]

Solution:

5 vol of oxygen are used to burn = 1 vol of propane
or 5cm³ of oxygen are used to burn = 1cm³ of propane
1cm³ of oxygen used to burn = $\frac{1}{5}$ cm³ of propane
100cm³ of oxygen are used to burn = $\frac{1}{5} \times 100$ cm³ of propane
= 20cm³ of propane

Q7

Liquid water can be decomposed into hydrogen and oxygen under suitable conditions (e.g., electrolysis) via the balanced reaction:

\[2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)} \nonumber \]

If an experiment generates 2500 cm³ of $H_2$ gas, what volume of $O_2$ is generated under the same conditions of temperature and pressure?

Solution:
Gay Lussac's Law is applicable to substances in the gaseous state. In the given example, Gay Lussac's Law is applicable to H₂ and O₂ but not to H₂O.

2H₂O(l) $\rightarrow$ 2H₂(g) + O₂(g)

Oxygen liberated when hydrogen liberated is 2vol = 1vol
Oxygen liberated when hydrogen liberated is 2cm³ = 1cm³
Oxygen liberated when hydrogen liberated is 2500cm³ = $\frac{1}{2} \times 2500$ = 1250cm³

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