Question

An excess of chlorine gas, Cl 2, is passed through 60 cm3 of cold aqueous 0.1 mol dm–3 sodium
hydroxide. In a separate experiment an excess of chlorine gas is passed through 60 cm3 of hot
aqueous 0.1 mol dm–3 sodium hydroxide until no further reaction takes place.
How much more sodium chloride will be produced by the reaction with hot NaOH than with cold
NaOH?
A 0.002 moles
B 0.003 moles
C 0.005 moles
D 0.006 moles

Answer 1

Answer:

b number is a your answer

Answer 2

Answer:

A

Explanation:

First calculate the moles of NaOH : 60/1000 * 0.1

Moles of NaOH are 6 * 10^-3

Then write both the equations to check the molar ratio of NaOH and NaCl

• In cold dilute conditions, it is 2 : 1
• In hot concentrated conditions, it is 6 : 5

{ Cl2 + cold NaOH --> NaClO + 2NaCl + H2O

3Cl2 + hot 6NaOH --> NaClO3 + 5NaCl + 3H2O}

From 2 : 1 we deduce moles of NaCl are

3 * 10^-3

From 5 : 6 we deduce moles of NaCl are

5 * 10^-3

In the question they are asking

How many more moles of NaCl

Hence subtract the two and you will get your answer that is

2 * 10^-3