In iron sulphide, what is the ratio of iron(2) and sulphur?

Open in App

Solution

In iron sulphide ( FeS):
Mass of Fe = 56u
Mass of S = 32u
Ratio of Fe and S = Mass of Fe / Mass of S
= 56 / 32
= 7 / 4
So, ratio by mass of Fe : S = 7 : 4

Suggest Corrections

Similar questions

In the formation of iron sulphide, by heating iron and sulphur, the mass of iron(II) sulphide __________ when compared with a total mass of iron and sulphur.

0.7 g of iron reacts directly with 0.4 g of sulphur to form ferrous sulphide. If 2.8 g f iron dissolves in dilute HCl and excess of sodium sulphide solution is added, 4.4 g of iron sulphide is precipitated. If it obeys the law of constant composition, what is the ratio of weights of Fe : S ?

Iron and sulphur when ___ together react to produce iron sulphide.

0.7 g of iron reacts directly with 0.4 g of sulphur to form ferrous sulphide. If 2.8 g of iron dissolves in dilute HCl and excess of sodium sulphide solution is added, 4.4 g of iron sulphide is precipitated. If it obeys the law of constant composition, what is the ratio of weights of Fe : S ?

Q. How can you form iron sulphide from a mixture of iron and sulphur?

Join BYJU'S Learning Program

In iron sulphide, what is the ratio of iron(2) and sulphur?

In iron sulphide ( FeS): Mass of Fe = 56uMass of S = 32uRatio of Fe and S = Mass of Fe / Mass of S= 56 / 32= 7 / 4So, ratio by mass of Fe : S = 7 : 4

In iron sulphide ( FeS):
Mass of Fe = 56u
Mass of S = 32u
Ratio of Fe and S = Mass of Fe / Mass of S
= 56 / 32
= 7 / 4
So, ratio by mass of Fe : S = 7 : 4