Question

$1$ lit of oxygen and $3$ lit of $S{O_2}$ at STP are reacted to produce sulphur trioxide. What would be the ratio between volume of sulphur trioxide and that of sulphur dioxide after reaction and weight of $S{O_3}$ formed (in grams) respectively?
$1:2;\;7.14\;gms$
$2:1;\;14.28\;gms$
$2:1;\;7.14\;gms$
$1:1;\;7.14\;gms$

Answer 1

As we know that sulphur dioxide reacts with oxygen to form sulphur trioxide and the volume at standard temperature and pressure conditions is equivalent to the ratio mass of the given solute to the molecular mass of the solute which is multiplied by $22.4\;L$ that is the standard volume at STP.

Complete solution:
As we know that two molecules of sulphur dioxide react with one molecule of oxygen to result in the formation of two molecules in sulphur trioxide. We can show this balanced equation through the given below reaction:
$2S{O_2} + {O_2} \to 2S{O_3}$
So, we can say that two litres of sulphur dioxide reacts with two litres of sulphur trioxide. We can also say that one litres of oxygen reacts with two litres of sulphur trioxide.
We are given that one litres of oxygen and three litres of sulphur dioxide reacts to form sulphur trioxide. So we can say that three litres of sulphur trioxide is formed.
Thus the ratio between the volumes of sulphur trioxide and that of the sulphur dioxide will be:
$\dfrac{{{V_{S{O_3}}}}}{{{V_{S{O_2}}}}} = \dfrac{3}{3}$
So, the ratio of volumes is $1:1$.
Now, to calculate the weight sulphur trioxide in grams we can relate its moles and volumes using the formula: $moles = \dfrac{{mass}}{{molecular\;mass}} = \dfrac{{volume\;at\;STP}}{{22.4L}}$
We can calculate the mass of sulphur trioxide having molecular mass $80.06\;g$ which would be:
$mass = \dfrac{{2 \times 80.06}}{{22.4}}$
$mass = 7.14\;gms$
Therefore, from the above explanation we can say that the correct option is (D).

Note: Always remember that balancing the equation informs us about the parameters which are fixed and the parameters which we need to calculate. So the first step in calculating the volume and mass should be the balancing of the given reaction. Hence, in the above reaction, two sulphur dioxide reacts with one oxygen molecule to give two sulphur trioxide molecules.