How many grams of calcium sulfite are produced daily by this power plant?

S

O

2

(

g

)

+

C

a

O

(

s

)

→

C

a

S

O

3

(

s

)

1 US ton

=

907.19

⋅

k

g

Mass of coal (kg)

=

2000

⋅

ton

*

907

⋅

k

g

⋅

ton

-

1

=

1.814

*

10

6

⋅

k

g

.

Mass of sulfur

=

1.814

*

10

6

⋅

k

g

*

2.5

%

=

45

,

350

⋅

k

g

Moles of sulfur

=

45

,

350

⋅

k

g

*

10

3

⋅

g

⋅

k

g

-

1

32.06

⋅

g

⋅

m

o

l

-

1

=

1.41

*

10

6

⋅

m

o

l

.

And thus we may have

1.41

*

10

6

⋅

m

o

l

S

O

2

produced during the daily operation.

Given the equivalence of sulfur dioxide and calcium oxide in the first equation, clearly we need

1.41

*

10

6

⋅

m

o

l

C

a

O

to produce

1.41

*

10

6

⋅

m

o

l

C

a

S

O

3

.

Mass of calcium sulfite produced daily

=

1.41

*

10

6

⋅

m

o

l

*

120.17

⋅

g

⋅

m

o

l

-

1

=

1.70

*

10

8

⋅

g

=

1.70

*

10

5

⋅

k

g

.