Question

At constant temperature and pressure, 5.0 L of SO2 is combined with 3.0 L of O2 according to the equation:

Answer 1

The balanced chemical reaction is:

2SO2 + O2=2SO3

We are given the amounts of the reactants to be consumed in the reaction. These amounts are the starting point for the calculations.

5.0 L SO2 ( 1 mol / 22.4 L ) = 0.2232 mol SO2
3.0 L O2  ( 1 mol / 22.4 L ) = 0.1339 mol O2

From the chemical reaction, the mole ratio of the reactants 2:1. Therefore, the limiting reactant is SO2.

0.2232 mol SO2 ( 1 mol SO3 / 2 mol SO2) ( 22.4 L / 1 mol ) = 2.4998 L SO3

Answer 2

Answer : The volume of mixture is, 4.99 liters.

Explanation :

First we have to calculate the moles of $SO_2$ and $O_2$.

As we know that at STP, 1 mole of substance contains 22.4 L volume of substance.

As, 22.4 L volume of $SO_2$ present in 1 mole of $SO_2$

So, 5 L volume of $SO_2$ present in $\frac{5}{22.4}=0.223$ mole of $SO_2$

And,

As, 22.4 L volume of $O_2$ present in 1 mole of $O_2$

So, 3 L volume of $O_2$ present in $\frac{3}{22.4}=0.134$ mole of $O_2$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2SO_2+O_2\rightarrow 2SO_3$

From the balanced reaction we conclude that

As, 2 mole of $SO_2$ react with 1 mole of $O_2$

So, 0.223 moles of $SO_2$ react with $\frac{0.223}{2}=0.1115$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $SO_3$

From the reaction, we conclude that

As, 2 mole of $SO_2$ react to give 2 mole of $SO_3$

So, 0.223 moles of $SO_2$ react to give 0.223 moles of $SO_3$

Now we have to calculate the volume of $SO_3$.

As, 1 mole of $SO_3$ gas occupy 22.4 L volume of $SO_3$

So, 0.223 mole of $SO_3$ gas occupy $0.223\times 22.4=4.99L$ volume of $SO_3$

Therefore, the volume of the mixture is 4.99 liters.