Question

If 10.0g of powdered iron is heated with 10.0g of sulfur in an open crucible, what is the mass of iron (II) sulfide that is formed? What mass of excess reactant is leftover? The reaction is as follows: Fe+S8–> FeS

Answer 1

See Explanation

Step-by-step explanation:

8Fe + S₈ => 8FeS

Given: 10g 10g

moles 10g/56g/mol 10g/256g/mol

= 0.179mol Fe = 0.039mol S₈

Reduce => divide mole values by respective coefficients; smaller value is Limiting Reactant.

0.179/8 = 0.022 0.039/1 = 0.039

=> Fe is limiting reactant

8Fe + S₈ => 8FeS

Given: 10g/56g/mol 10g/256g/mol

= 0.179mol = 0.039 mol 0.179mol FeS produced

1/8(0.179)mol S₈ used (coefficients are equal,

= 0.022 mol S₈ used => moles Fe = moles FeS)

= (0.039 - 0.022)mol S₈ = 0.179mol FeS

remains in excess =(0.179mol)(88g/mol)

= 0.0166 mol S₈ (excess) = 15.8 g FeS

= (0.0166mol)(256g/mol) (Theoretical Yield)

= 4.26g S₈ in excess