Answer :
Final answer:
The amount of arsenic pentasulphide obtained when 35.5 g arsenic acid is treated with excess H2S in the presence of conc. HCl is 60.5 g.
Explanation:
The reaction given is:
As2O5 + 5H2S -> 2As2S5 + 5H2O
From the balanced equation, we can see that 1 mol of As2O5 reacts with 5 mol of H2S to produce 2 mol of As2S5. Therefore, the molar ratio between As2O5 and As2S5 is 1:2. To find the amount of As2S5 produced, we need to calculate the moles of As2O5 first:
Moles of As2O5 = mass / molar mass = 35.5 g / (2 x 2 x 16 + 5 x 32) g/mol = 35.5 g / 296 g/mol = 0.12 mol
Since the molar ratio between As2O5 and As2S5 is 1:2, the moles of As2S5 produced will be twice the moles of As2O5:
Moles of As2S5 = 2 x 0.12 mol = 0.24 mol
To calculate the mass of As2S5 produced, we can use the formula:
Mass of As2S5 = moles x molar mass = 0.24 mol x (2 x 2 x 32 + 5 x 32) g/mol = 0.24 mol x 252 g/mol = 60.5 g
Therefore, the amount of arsenic pentasulphide obtained is 60.5 g. None of the given options match this value, so none of the provided options are correct.
