Question

3.0 dm3 of sulfur dioxide is reacted with 2.0 dm3 of oxygen according to the equation below. 2SO2(g) + O2(g) → 2SO3(g) What volume of sulfur trioxide (in dm3) is formed? (Assume the reaction goes to completion and all gases are measured at the same temperature and pressure.)
A. 5.0
B. 4.0
C. 3.0
D. 2.0

Answer 1

Answer:

Consider,

2

S

O

2

(

g

)

+

O

2

(

g

)

→

2

S

O

3

(

g

)

Given,

2.0

dm

3

⋅

1.43

g

dm

3

⋅

mol

32

g

≈

8.94

⋅

10

−

2

mol

of

O

2

, and

3.0

dm

3

⋅

2.62

g

dm

3

⋅

mol

64.1

g

≈

0.123

mol

of

S

O

2

Alternatively you could use the ideal gas law to derive the moles, but I used density because I could look them up (on an exam the former is probably your method of choice).

From the preceding data, we can conclude that

S

O

2

is the limiting reactant. We need twice as much, according to the stoichiometry, and dividing it by two gives us

n

S

O

2

<

n

O

2

. Hence,

0.123

mol

⋅

2

S

O

3

2

S

O

2

⋅

80.1

g

mol

⋅

dm

3

2.62

g

≈

3.76

dm

3

of

S

O

3

Explanation:

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