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.
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
To determine if your calculation is correct, let's go through the steps together.
1. First, we calculate the number of moles of zinc (Zn) and sulfur (S8) using their respective molar masses.
- Moles of Zn = 6.2 g / 65.39 g/mol = 0.095 mol
- Moles of S8 = 4.5 g / 256.48 g/mol = 0.018 mol
2. Next, we need to determine the limiting reactant, which is the reactant that will be entirely consumed and determines the amount of product formed. To do this, we compare the mole ratios between Zn and S8 in the balanced equation.
From the equation: 8Zn + S8 >> 8ZnS
The ratio of moles of Zn to moles of S8 is 8:1.
- If the ratio of moles Zn:moles S8 is less than 8:1, Zn is the limiting reactant.
- If the ratio of moles Zn:moles S8 is greater than 8:1, S8 is the limiting reactant.
Let's check the ratio:
0.095 mol Zn : 0.018 mol S8
Since the ratio is less than 8:1, Zn is the limiting reactant.
3. Now, let's calculate the moles of ZnS formed using the limiting reactant.
Using the limiting reactant (Zn), we use the stoichiometry from the balanced equation to find the moles of ZnS formed:
- Moles of ZnS = 0.095 mol Zn x (8 mol ZnS / 8 mol Zn) = 0.095 mol ZnS
4. Finally, we calculate the mass of ZnS using its molar mass.
Molar mass of ZnS = 97.47 g/mol
Mass of ZnS = 0.095 mol x 97.47 g/mol = 9.19 g
So, the correct mass of zinc sulphide (ZnS) produced when 6.2 g of zinc and 4.5 g of sulfur are reacted is 9.19 grams.