Question

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.

Answer 1

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.

Answer 2

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?

Answer 3

Hannah, that's my error. Zn is the limiting reagent, not S and the correct grams ZnS formed is 9.24.

Answer 4

Thank you.

Answer 5

12g of Zn reacted with 6.5g of sulphur to form zinc sulphide, how many grams of ZinS can be formed from this particular reaction