How many moles of au2s3 is required to form 56 grams of h2s at stp?

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

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally