I don't know how they got that, there could have been a number of ways. One way is to find the percent of Au in Au2S
percentAu= atomicmassAu/molemassAu2S
Then, 1017moleAu/percentAu * atomicmassAu/1 mole
How many grams of Au2S can be obtained from 1017 mol of Au? The answer is 249. How did they get that?
2 answers
I must admit that I didn't understand the question and had to work it backwards.
Start from a balanced equation
2Au + S -> Au2S
if we are starting with 1.17 mol Au (rather than 1017 mol Au which is rather a lot!)
then as 2 moles of Au give 1 mole of Au2S from the equation, 1.17 mol Au will yield 0.585 mol Au2S.
The molar mass for Au2S is 426 g mol^-1, then the mass obtained is
0.585 mol^-1 x 426 g mol^-1 = 249 g
Start from a balanced equation
2Au + S -> Au2S
if we are starting with 1.17 mol Au (rather than 1017 mol Au which is rather a lot!)
then as 2 moles of Au give 1 mole of Au2S from the equation, 1.17 mol Au will yield 0.585 mol Au2S.
The molar mass for Au2S is 426 g mol^-1, then the mass obtained is
0.585 mol^-1 x 426 g mol^-1 = 249 g
1 answer