Question

17. Sulphuric acid reacts with sodium hydroxide as follows:
HSO4 + 2NaOH → Na2SO4 + 2H2O
When 1 L of 0.1 M sulphuric acid solution is allowed to
react with 1 L of 0.1 M sodium hydroxide solution, the
amount of sodium sulphate formed and its molarity in the
solution obtained is :
(a) 0.1 mol L-1
(b) 7.10 g
(c) 0.025 mol L -1
(d) 3.55 g
29​

Answer 1

Answer:

Hey!! The answer is (b) 7.10

Explanation:

Moles of H2S0

4

taken = 0.1 moles

Moles of NaOH taken = 0.1 moles

As H2S0

4

and NaOH react in ratio 1:2, so 0.1 moles of H2S0

4

reacts with 0.2 mole of NaOH which we don’t have.

0.1 mole of NaOH reacts with 0.05 mole of H2S0

4

, so NaOH is Limiting Reactant. Product is calculated w.r.t limiting reactant so Number of moles of Na

2

S0

4

formed will also be equal to 0.05.

Mass of Na

2

S0

4

=0.05×142=7.1g