Question

If the mass of $S{O_2}$ used in the reaction $2S{O_2} + {O_2} \to S{O_3}$ is 2g then the mass of $S{O_3}$ produced is:
A. 2.5g
B. 2g
C. 1g
D. 1.5g

Answer 1

First we have to determine the mole ratio between $S{O_2}$ and $S{O_3}$. We have to convert the mass of sulfur dioxide in grams to mass of sulfur dioxide in moles using molar mass of sulfur dioxide. We have to then calculate the mass of sulfur trioxide using the moles of sulfur dioxide and molar mass of sulfur trioxide.

Complete step by step answer:
Given data contains,
Mass of sulfur dioxide is 2g.
The given chemical equation is,
$2S{O_2} + {O_2} \to 2S{O_3}$
In the given reaction two moles of sulfur dioxide are reacted with one moles of oxygen to form two moles of sulfur trioxide. The mole ratio between sulfur dioxide and sulfur trioxide is $2:2 = 1$. So, we have found the mole ratio between sulfur dioxide and sulfur trioxide as 1.
Let us now convert the two grams of sulfur dioxide into moles.
We know that the molar mass of sulfur dioxide is $64g/mol$. When we divide the mass of sulfur dioxide by the molar mass of sulfur dioxide, we get the moles of sulfur dioxide.
$Moles = \dfrac{{Grams}}{{{\text{Molar mass}}}}$
Let us now substitute the grams of sulfur dioxide and molar mass of sulfur dioxide to get the moles of sulfur dioxide.
$Moles = \dfrac{{2g}}{{64g/mol}}$
$Moles = 0.03125mol$
We have calculated the moles of sulfur dioxide as $0.03125mol$.
Since, the mole ratio between sulfur dioxide and sulfur trioxide as 1, the moles of sulfur trioxide formed would also be $0.03125mol$.
So, from the moles of sulfur trioxide, let us now calculate the mass of sulfur trioxide using the molar mass of sulfur trioxide.
We know that molar mass of sulfur trioxide is $80g/mol$.
When we multiply the moles of sulfur dioxide by the molar mass of sulfur trioxide, we get the mass of sulfur trioxide produced.
$Grams = Moles \times {\text{Molar mass}}$
Let us now substitute the values of moles and molar mass of sulfur trioxide.
$Grams = Moles \times {\text{Molar mass}}$
Substituting the known values we get,
$\Rightarrow Grams = 0.03125mol \times 80g/mol$
On simplifying we get,
$\Rightarrow Grams = 2.5g$
The mass of sulfur trioxide produced is $2.5g$.

So, the correct answer is Option A.

Note: While determining the mole ratio between the reactant and the product, it is mandatory to check if the reaction is balanced (or) unbalanced. If the reaction is unbalanced, we have to balance the reaction. If the reaction is unbalanced, there would be error while determining the ratio, which could also error in mass calculations.