Question

The minimum quantity in grams of H2s needed to precipitate 75 gm As+3 will be neatly

Answer 1

Reaction for the given condition will be ⇒

Cu²⁺ + H₂S ------------→ Cu₂S + 2[H]⁺

In this Reaction, 

1 mole of the Cu²⁺ reacts with the one mole of the H₂S.

Molar Mass of the H₂S = 2 + 32 = 34 g. 

∴ 63.5 g of the Cu²⁺ reacts with the 34 g of the H₂S.

Hence, the mass of the H₂S needed in the grams is 34 g.

Hope it helps.

From using this method you have your  answer

Answer 2

Reaction for the given condition will be ⇒

Cu²⁺ + H₂S ------------→ Cu₂S + 2[H]⁺

In this Reaction, 

1 mole of the Cu²⁺ reacts with the one mole of the H₂S.

Molar Mass of the H₂S = 2 + 32 = 34 g. 

∴ 63.5 g of the Cu²⁺ reacts with the 34 g of the H₂S.

Hence, the mass of the H₂S needed in the grams is 34 g.

Hope it helps.