1. what is the mass of the iron (fe) involved in this reaction? (4 pts) s + fe → fes 32 grams 88 grams

This question is incomplete. We would need a starting amount for one of the products, either in mass or moles.

We can complete the question by saying How many grams of FeS are produced if 10.0 grams of iron are combined with an excess of sulfur.

We begin with a balanced chemical equation that is provided in the question.

8

F

e

+

S

8

→

8

F

e

S

Next we determine what we have and what we want.

We have 10.0 grams of iron and we want grams of iron(II) sulfide.

We set up a roadmap to solve the problem

g

r

a

m

s

F

e

→

m

o

l

F

e

→

m

o

l

F

e

S

→

g

r

a

m

s

F

e

S

We need the molar mass (

g

f

m

) of

F

e

and

F

e

S

.

F

e

= 55.8 g/mol and

F

e

S

= 87.8 g/mol (55.8 +32.0).

We need the mole ratio between

F

e

:

F

e

S

8:8. This comes from the coefficients of the balanced chemical equation.

Now we set up conversion factors following the roadmap from above.

Unit we want in the numerator, unit to cancel in the denominator.

10.0

g

F

e

x

1

m

o

l

F

e

55.8

g

F

e

x

8

m

o

l

F

e

S

8

m

o

l

F

e

x

87.8

g

F

e

S

1

m

o

l

F

e

S

=

Multiply the numerators, and divide the denominators.

The outcome is

15.7

g

F

e

S

will be produced from 10.0 grams

the correct answer is:

a. formation of precipitation