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
Answers
Answer:
I hope it was helpful :)
Explanation:
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
would ideally be produced.
Given: volume of sulfur dioxide = 3.0 dm³
volume of oxygen = 2.0 dm³
To find: volume of sulfur trioxide
Solution: The reaction is -
2SO₂ + O₂ → 2SO₃
According to the law of conservation of mass,
Two moles of sulfur dioxide reacts with one mole of oxygen gives two moles of sulfur trioxide.
So, 3 dm³ of sulfur dioxide reacts with 1.5 dm³ to form 3 dm³ of sulfur trioxide.
Therefore, 3 dm³ of sulfur trioxide is formed, so the correct option is (c).
- The law of conservation of mass governs the balancing of a chemical equation. According to this law, the total number of moles of reactant is equal to the total number of moles of products.
- If the number of moles on the reactant side is not equal to the number of moles on the product side, then the reaction is said to be an unbalanced reaction.