Chemistry  Please Check!
posted by Hannah on .
Determine the mass of zinc sulphide, ZnS, produced when 6.2g of zinc and 4.5g of sulphur are reacted.
Equation given: Zn + S8 >> ZnS
Balanced equation: 8Zn + S8 >> 8ZnS
nZn= 6.2 g/ (65.39 g/mol) = 0.095 mol
nS8= 4.5 g/(256.48 g/mol) = 0.018 mol
nS8 = 0.095 mol Zn x (1 mol S8/8 mol Zn) = 0.012 mol
0.095 mole Zn x (8 mole ZnS / 8 moles Zn) = 0.095 moles ZnS
mZnS = 0.095 mol x 97.47 g/mol
mZnS = 9.24 g
Therefore, the mass of zinc sulphide produced when 6.2 g of zinc and 4.5 g of sulphur are reacted, 9.24 g of sulphide are produced.
Is 9.24 g the right answer? As well as everything I've done to get to 9.24 grams? If you could let me know that would be great! Thanks.

I don't see any calculations with the S8. I think S8 is the limiting reagent. If that is true 9.24 g ZnS can't be right.

mZnS = 0.018 x 97.47 g/mol
mZnS = 1.71 g
I am pretty positive the first part of is right but, if my final answer is wrong again, could you explain the question for me? 
Hannah, that's my error. Zn is the limiting reagent, not S and the correct grams ZnS formed is 9.24.

Thank you.