Asked by Hannah

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.

Answers

Answered by DrBob222
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.
Answered by Hannah
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?
Answered by DrBob222
Hannah, that's my error. Zn is the limiting reagent, not S and the correct grams ZnS formed is 9.24.
Answered by Hannah
Thank you.
Answer
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
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