Question

The compound SF6 is made by burning sulfur in an atmosphere of fluorine. The balanced equation is S8 + 24 F2(g) → 8 SF6(g) Starting with a mixture of 724. g of sulfur, S8, and 2670 g of F2,

What is the limiting reagent?

What amount (in grams) of SF6 could be produced?

What amount (in grams) of the excess reagent will remain?

Answer 1

S8    +    24 F2(g) → 8 SF6(g)

1 mole    24 mole

no of moles of S8 = 724/256.5 = 2.82 moles

no of mole of F2   = 2670/38   = 70.2 moles

from balanced equation

1 mole of S8 react with 24 moles of F2

2.82 moles of S8 react with = 24*2.82/1 = 67.68 moles are required

limiting reagent is S8

1 mole of S8 react with F2 to from 8 moles of SF6

2.82 moles of S8 react with F2 to from = 2.82*8 = 22.56 moles of SF6

mass of SF6 = no of moles of SF6* molar mass

                   = 22.56*146 = 3293.76gm of SF6

Excess reagent is F2 = 70.2-67.68 =2.52 moles

mass of F2 = no of moles * molar mass

                = 2.52*38 = 95.76gm of F2 >>>> answer