Question

Determine the mass of 21.05 moles of iron(III) sulfide.

Answer 1

The mass of 21.05 moles of iron(III) sulfide is 4,372.085 grams.

Step-by-step explanation:

Iron (III) sulfide is a chemical compound with the formula Fe₂S₃.

Being the molar mass of the elements:

• Fe: 55.85 g/mole
• S: 32 g/mole

then the molar mass of iron sulfide (III) is:

Fe₂S₃= 2* 55.85 g/mole + 3* 32 g/mole= 207.7 g/mole

Molar mass is the amount of mass that a substance contains in one mole. Then the rule of three can be applied: if in 1 mole of iron (III) sulfide there are 207.7 grams, in 21.05 moles how much mass is there?

`mass=(21.05 moles*207.7 grams)/(1 mole)`

mass= 4,372.085 grams

The mass of 21.05 moles of iron(III) sulfide is 4,372.085 grams.