Question

when 1 L of 0.1 M sulphuric acid solution is allowed to react with 1L of 0.1M of sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution is obtained is

Answer 1

(b, c) Moles of H2S04 taken = 0.1 moles

Moles of NaOH taken = 0.1 moles

As H2S04 and NaOH react in ratio 1:2, so 0.1 moles of H2S04 reacts with  0.2 mole of NaOH which we don’t have.

0.1 mole of NaOH reacts with 0.05 mole of H2S04, so NaOH is Limiting reactant. Product is calculated w.r.t limiting reactant so Number of moles of Na2S04 formed will also be equal to 0.05.

Mass of Na2S04 = 0.05 x 142 = 7.1 g

Answer 2

Answer:

(a)7.1 gram

(b)0.025 mol/ l

Explanation:

plz refer to the attachment ☺️