What mass of Au is produced when 0.05 mol of Au2S3 is reduced completely with excess H2?

Question

Can anyone tell me how to solve this problem? I don't need the answer I just need to know how to do it... like should we convert from grams to mols? and then what? I'm really not sure how to even start out the problem ... please help!!! Thank you!

Answer 1

you write the reaction

Au2S3 +3H2---> 2 Au +3SH2

you see that 1 mole ofAu2S3 correspond to 2 moles of Au

0.05mole of Au2S3 correspond to 0.05*2=0.1mole of Au

as the mplecular weight of Au =197

0.1 mole =19.7g the answer

Answer 2

if you have .05 mol of Au2S3 in the reaction(not balanced):

Au2S3 + H2 --> 2Au +H2S

meaning for every 1 Au2S3 molecule that you have you will get 2 Au molecules.

so .05 mols will turn into 0.1 mol Au. then you multiply .1 by the atomic mass of gold (the number under the symbol on the periodic table) and you should get the mass of Au produced:

19.7 g

Answer 3

1) First, we need the reaction:

Au(3+) + 3e(-) -----> Au(0)

2) Second, you can compute the percentage of gold in this compound (divide atomic mass of 2Au by molecular mass of Au2S3)

3) Knowing the % of Au multiply it by 0.05 moles. In this moment you know the moles of Au.

4) Multiply the obtained moles in the previous step and you will get the mass of reduced gold.

That's it!
Good luck!