You have been given the following balanced chemical equation: 4 Au + 3 O2 → 2 Au2O3. How many moles of O2 are needed to use up 6.00 g Au?

Answer 1

Hi!

This stoichiometry question has you going from grams of Au to mol of O₂. You have to be able to cancel out units in order to change from grams to mol. In order to solve this problem, you need to first identify the ratio of O₂ to Au (this comes from your coefficients in the balanced equation). The ratio is 3:4. For every 3 mol of O₂ you have 4 mol of Au.

Next, you need to identify the molar mass of Au (in this case it is just the atomic mass of Au) which is 196.97g = 1 mol Au.

Start with what you are given: 6.00 g Au

Then you will need to cancel g Au. To do this the units need to be diagonal from each other. Here is the first part of this problem:

6.00 g Au x 1 mol Au

1 196.97 g Au

Now that g Au cancels, you will need to cancel mol Au. Again, units should be diagnol. This is where you will use the ratio you identified above.

6.00 g Au x 1 mol Au x 3 mol O₂

1 196.97g Au 4 mol Au

Now solve the problem in same way you would any fraction: 6 x 1 x 3 = 18

1 x 196.97 x 4 = 787.88

Top divided by bottom: 18/787.88 = 0.0228 mol O₂

I hope this helps!