Question

Equal masses of iron and sulpher are heat together to from FeS. What fration of the original mass of excess reactant is left unreacted.(Fe=56,S=32) ​

Answer 1

hey mate here is your ans if your qun

Answer:

Let equal mass = w

Mole of Fe = w/56

Mole of sulphur = w/32

Now it should form mole of FeS = w/88 = .0113 w

Now w/56 = 0.0178 w

Left = 0.0178 - 0.0113 = 0.0065

Now reacted = 0.0065 * 88/100 = .572

Hence unreacted = 1 - 0.572 = 0.428 = 0.43.....

hope it helps u

Answer 2

In this reaction, the sulfur melts and reacts with the iron exothermically to form iron(II) sulfide, which contains equal amounts of iron and sulfur. So let their equal mass = w

Then, Mole of Fe = w/56 [given]

And, Mole of sulphur = w/32 [given]

Now it should form mole of FeS = w/88 = .0113 w

Now w/56 = 0.0178 w

The Left = 0.0178 - 0.0113 = 0.0065

Now reacted = 0.0065 * 88/100 = .572

Hence unreacted = 1 - 0.572 = 0.428 = 0.43