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