Question

How many moles of Au2S3 is required to form 56 grams of H2S at STP?

Answer 1

Answer:

0.55 mol Au₂S₃

Explanation:

Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.

1. Gather all the information in one place:

M_r:                          34.08

          Au₂S₃ + … ⟶ 3H₂S + …  

m/g:                             56  

2. Calculate the moles of H₂S

Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)

                      = 1.64 mol H₂S

3. Calculate the moles of Au₂S₃

The molar ratio is 1 mol Au₂S₃/3 mol H₂S.

Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)

                         = 0.55 mol Au₂S₃