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Question

6.0 dm3 of sulfur dioxide is reacted with 4.0 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.

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Solution

Dear Student

Since, gases are measured at same temperature and pressure, therefore, 2 dm3 of sulphur dioxide, SO2 reacts with 1 dm3 of oxygen, O2 to produce 2 dm3 of sulphur trioxide, SO3, that is -
2SO2 + O2 -------> 2SO3
2 L 1 L 2 L (because, 1 dm3 = 1 L)

Now, ​6 dm3 of sulphur dioxide, SO2 would react with 4 dm3 of oxygen, O2 to produce x dm3 of sulphur trioxide, SO3, that is -
2SO2 + O2 -------> 2SO3
6 L 4 L x L

Therefore, substituing the values, we get -

SO2 (2 L) produces SO3 (2 L)
So,
SO2 (6 L) should produce SO3 (6 L)

Thus, 6 L of SO3 or sulphur trioxide should be formed.

Regards

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