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Mole Calculations.

A mixture containing 2.8 g of iron and 2.0 g of sulphur is heated together. What mass of iron(II) sulphide is produced ? Answer is 4.4g. Not sure why that is the answer.
Reply 1
Original post by Paranoid_Glitch
A mixture containing 2.8 g of iron and 2.0 g of sulphur is heated together. What mass of iron(II) sulphide is produced ? Answer is 4.4g. Not sure why that is the answer.


What did you think was the answer? And why?
Original post by Ruineth
What did you think was the answer? And why?
4.4grams. Got the moles of iron and multiplied it by the relative formula mass of Fes, & got the mass to be 4.4g. Got the moles of sulphur and multiplied it by the relative formula mass of Fes got 5.5g. Based on mass conservation the maximum theoretical yield can only be 4.8g. So i gave the answer as 4.4g not sure why.
Think about the overall equation and look at the molar ratio (1 to 1 to 1). Work out the number of moles of both Fe and S you'll find that sulphur is in excess so iron is the limiting reactant and reacts completely whereas some sulphur will remain at the end of the reaction - so the maximum number of moles of FeS that you could possibly form from this is equal to the number of moles of iron. You multiply this by the molar mass of FeS and get 4.4g.
(edited 7 years ago)
Original post by victoria98
Think about the overall equation and look at the molar ratio. Work out the number of moles of both Fe and S you'll find that sulphur is in excess so iron is the limiting reactant and reacts completely whereas some sulphur will remain at the end of the reaction - so the maximum number of moles of FeS that you could possibly form from this is equal to the number of moles of iron. You multiply this by the molar mass of FeS and get 4.4g.
How would i be able to tell what is in excess and hence what is the limiting reagent?
Reply 5
Original post by Paranoid_Glitch
How would i be able to tell what is in excess and hence what is the limiting reagent?


Work out the number of moles in each of 2.8g of Fe and 2.0g of S.
The one with the more moles would be in excess, the other would then be the limiting reagent
Original post by Paranoid_Glitch
How would i be able to tell what is in excess and hence what is the limiting reagent?


Fe + S -> FeS - you can see that it's a 1: 1: 1 molar ratio.

Number of moles of Fe = 2.8/55.8 = 0.0502 mol
Number of moles of S =2/32.1 = 0.0623 mol - EXCESS

From this you can (hopefully) see that, since the ratio Fe : S is 1 to 1, all of the iron will react with sulphur but some sulphur will remain because you have more of it so there's no more iron to magically react with it. Therefore iron is the limiting reagent.

0.0502 moles is the largest amount of FeS that you can form. This multiplied by the molar mass = 4.4g.
Original post by victoria98
Fe + S -> FeS - you can see that it's a 1: 1: 1 molar ratio.

Number of moles of Fe = 2.8/55.8 = 0.0502 mol
Number of moles of S =2/32.1 = 0.0623 mol - EXCESS

From this you can (hopefully) see that, since the ratio Fe : S is 1 to 1, all of the iron will react with sulphur but some sulphur will remain because you have more of it so there's no more iron to magically react with it. Therefore iron is the limiting reagent.

0.0502 moles is the largest amount of FeS that you can form. This multiplied by the molar mass = 4.4g.

Oh thanks :h:. If the number of moles of each of the substances reacting were equal would not product be formed?

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