Question

What is maximum amount of 100% pure H2SO4 can be obtained by addition of H2O in 100 g of 109% oleum

Answer 1

Answer:

The additional mass comes form amount of H2O required to convert all SO3 present in 100 g oleum into complete H2SO4. Reaction involved is SO3 + H2O = H2SO4. Here, 18 g H2O is required for 80 g SO3. So, in 109 % oleum, 9 g H2O would be required for 40 g SO3.

Answer 2

Given:

Mass of oleum = 100 gm

Label of Oleum = 109 %

To Find:

Maximum amount of 100% pure H2SO4 that can be obtained from the sample by addition of water.

Calculation:

- Oleum (H2S2O7) is a mixture of H2SO4 and SO3.

- The reaction of oleum with water is given as:

H2S2O7 + H2O → 2 H2SO4

- 109% of oleum implies that when 100g of oleum is treated with water, then 109 gm of H2SO4 is obtained.

- Hence the amount of H2SO4 produced by the addition of water is 109 gm.