Calculate Mass of Au Produced from 0.0500 mol Au2S3

[Quincy]

Homework Statement

What mass of Au is produced when 0.0500 mol of Au₂S₃ is reduced completely with excess H₂?

Homework Equations

The Attempt at a Solution

.100 mol Au³⁺ x 197/1 = 19.7 g

19.7 g is apparently wrong though according to the book; what am I doing wrong?

 

[Borek]

Look for a new book, 19.7g is the correct answer.

 

[Blueshift5]

To calculate the mass of Au produced from 0.0500 mol of Au2S3, we first need to write a balanced chemical equation for the reaction between Au2S3 and H2. Based on the given information, the equation would be:

Au2S3 + 3H2 → 2Au + 3H2S

From this equation, we can see that for every 1 mole of Au2S3, 2 moles of Au will be produced. Therefore, for 0.0500 mol of Au2S3, we can expect to produce 0.100 mol of Au (0.0500 mol x 2 moles Au/1 mole Au2S3).

Now, to calculate the mass of Au produced, we can use the molar mass of Au, which is 197 g/mol. So, the mass of Au produced would be:

0.100 mol Au x 197 g/mol = 19.7 g Au

This is the same answer you got, so it seems like you did the calculation correctly. It's possible that the book has a different answer or that there is a typo. It's always a good idea to double check your calculations and make sure you are using the correct molar mass.