Question

If you have $2.0$ grams of $S{{O}_{2}}$ available to react with the oxygen, how many grams of $S{{O}_{3}}$ can be produced in the following reaction?
$2S{{O}_{2}}\to 2S{{O}_{3}}$

Answer 1

Hint: In this, first we will the reaction and find the molecular weight of Sulphur dioxide and Sulphur trioxide and we know the weight of the Sulphur dioxide, then by applying the unitary method, we can easily find out the weight of the Sulphur trioxide. Now solve it.

Complete answer:
- When oxygen is made to undergo reaction with the Sulphur dioxide, it undergoes the combination reaction and results in the formation of the compound, the Sulphur trioxide.
- Combination reactions are those reactions in which two or more reactants combine together and result in the formation of the one single product.
The reaction of the oxygen and Sulphur dioxide is supposed to take place as follows:
 $2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}$ ----------(1)
Now considering the statement as;
The molecular mass of Sulphur dioxide i.e. $S{{O}_{2}}=32+32=64g$
And the molecular mass of Sulphur trioxide i.e. $S{{O}_{3}}=32+48=80g$
From the equation (1), we comes to know that;
$64g$ of the Sulphur dioxide produces = $80g$ of the Sulphur trioxide
Then,
$1g$ of the Sulphur dioxide produces = $\dfrac{80}{64}g$ of the Sulphur trioxide
And
$2g$ of the Sulphur dioxide produces = $\dfrac{80}{64}\times 2=2.5g$ of the Sulphur trioxide
Thus, when we have $2.0$ grams of $S{{O}_{2}}$ available to react with the oxygen, $2.5$ grams of $S{{O}_{3}}$ is produced in the following reaction:
$2S{{O}_{2}}\to 2S{{O}_{3}}$ .

Note: We can also solve this statement by using the mole concept. In this, first we will find the moles of the Sulphur dioxide by using the molecular weight and given mass and applying the formula as: $mole=\dfrac{given\text{ }mass}{\text{ molecular mass}}$. Now we know that the moles of the Sulphur trioxide will be the same as that of the Sulphur dioxide, then by using the same formula, we can find the weight of the Sulphur trioxide.