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.
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?
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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