Q: A piece of sodium when dropped in
water catches fire, why? Explain thr
formation of Na,O by transfer of electron
Answers
Answer:
ok wait........
Explanation:
(i) When sodium metal is dropped in water, an exothermic reaction occurs due to which the evolved hydrogen gas catches fire.
2Na(s)+2H
2
O(l)→2NaOH(aq)+H
2
(ii) sodium metal is heated in free supply of air to form sodium peroxide and sodium oxide.
2Na(s)+O
2
(g)→Na
2
O(s) minor product
Na
2
O
2
(s)+2H
2
O(l)→2NaOH(aq)+H
2
O
2
(l)
(iii) sodium peroxide dissolves in water to form hydrogen peroxide.
Na
2
O
2
(s)+2H
2
O(l)→2NaOH(aq)+H
2
O
2
(l)
Answer:
(i) When sodium metal is dropped in water, an exothermic reaction occurs due to which the evolved hydrogen gas catches fire.
2Na(s)+2H
2
O(l)→2NaOH(aq)+H
2
(ii) sodium metal is heated in free supply of air to form sodium peroxide and sodium oxide.
2Na(s)+O
2
(g)→Na
2
O(s) minor product
Na
2
O
2
(s)+2H
2
O(l)→2NaOH(aq)+H
2
O
2
(l)
(iii) sodium peroxide dissolves in water to form hydrogen peroxide.
Na
2
O
2
(s)+2H
2
O(l)→2NaOH(aq)+H
2
O
2
(l)
Explanation:
(i) When sodium metal is dropped in water, an exothermic reaction occurs due to which the evolved hydrogen gas catches fire.
2Na(s)+2H
2
O(l)→2NaOH(aq)+H
2
(ii) sodium metal is heated in free supply of air to form sodium peroxide and sodium oxide.
2Na(s)+O
2
(g)→Na
2
O(s) minor product
Na
2
O
2
(s)+2H
2
O(l)→2NaOH(aq)+H
2
O
2
(l)
(iii) sodium peroxide dissolves in water to form hydrogen peroxide.
Na
2
O
2
(s)+2H
2
O(l)→2NaOH(aq)+H
2
O
2
(l)