(a) State Gay Lussac's law of combining volumes
(b) Hydrogen reacts with oxygen according to the following equation;
2H\(_{2(g)}\) + O\(_{2(g)}\) ---> 2H\(_2\)O\(_{(g)}\). If 50cm\(^3\) of hydrogen were sparked with 30cm\(^{3}\) of oxygen, calculate the volume of unused oxygen after cooling to the initial temperature and pressure.
(b) Hydrogen reacts with oxygen according to the following equation;
2H\(_{2(g)}\) + O\(_{2(g)}\) ---> 2H\(_2\)O\(_{(g)}\). If 50cm\(^3\) of hydrogen were sparked with 30cm\(^{3}\) of oxygen, calculate the volume of unused oxygen after cooling to the initial temperature and pressure.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) Gay - Lussac Law of combining volumes states that when gases react, they do so in volumes which are in simple ratios to one another and to the volume of the product, if gaseous, provided that the temperature and pressure remain constant.
(b) 2H\(_2\) + O\(_2\) -> 2H\(_2\)O. According to the equation, 2 volume of H\(_2\) requires 1 volume of oxygen.
50cm\(^3\) of H\(_2\) requires 25cm\(^3\) of oxygen,
volume of oxygen available = 30cm\(^3\),
volume of unused oxygen 30 - 25 = 5cm\(^3\).
(a) Gay - Lussac Law of combining volumes states that when gases react, they do so in volumes which are in simple ratios to one another and to the volume of the product, if gaseous, provided that the temperature and pressure remain constant.
(b) 2H\(_2\) + O\(_2\) -> 2H\(_2\)O. According to the equation, 2 volume of H\(_2\) requires 1 volume of oxygen.
50cm\(^3\) of H\(_2\) requires 25cm\(^3\) of oxygen,
volume of oxygen available = 30cm\(^3\),
volume of unused oxygen 30 - 25 = 5cm\(^3\).