Gunpowder spontaneously reacts when ignited. The reaction(s) can be :

2 K N O_{3} (s) + 4 C (s) \to K_{2} C O_{3} (s) + 3 C O (g) + N_{2} (g)

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2 K N O_{3} (s) + 2 S (s) \to K_{2} S O_{4} (s) + S O_{2} (g) + N_{2} (g)

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Both A & B

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None of the above

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Solution

The correct option is C Both A & B
Gunpowder is a mixture of sulphur, charcoal and potassium nitrate (nitre). Amongst these, potassium nitrate (saltpeter) is a major constituent of gun powder.
The following reactions takes place on ignition-
$2 K N O_{3} (s) + 2 S (s) \to K_{2} S O_{4} (s) + S O_{2} (g) + N_{2} (g)$
$2 K N O_{3} (s) + 4 C (s) \to K_{2} C O_{3} (s) + 3 C O (g) + N_{2} (g)$

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Similar questions

Q. How many atoms are present in the given reaction?

{P b ({N O}_{3})}_{2} + 2 K I \to P b I_{2} + 2 K N O_{3}

N_{2} (g) + {3 H}_{2} (g) ⇌ 2 {N H}_{3} (g) + h e a t

For the reaction above, what is the equilibrium expression?

N_{2} (g) + {3 H}_{2} (g) \to {2 N H}_{3} (g); Δ H^{O} = - 44 K c a l

For the above reaction, the correct statement is:

N_{2} (g)

reacts with

H_{2} (g)

in either of the following ways depending upon the supply of

H_{2} (g)

$N_{2} (g) + H_{2} (g) \to N_{2} H_{2} (l)$

$N_{2} (g) + 2 H_{2} (g) \to N_{2} H_{4} (g)$

If $5 L N_{2} (g)$ and $3 L H_{2} (g)$ are taken initially (at same temperature and pressure), then the contraction in volume after the reaction (in L) is:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g) + h e a t

For the reaction above, what is the effect on the equilibrium position when the pressure is decreased?

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