Respuesta :
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
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 : The volume of mixture is, 4.99 liters.
Explanation :
First we have to calculate the moles of [tex]SO_2[/tex] and [tex]O_2[/tex].
As we know that at STP, 1 mole of substance contains 22.4 L volume of substance.
As, 22.4 L volume of [tex]SO_2[/tex] present in 1 mole of [tex]SO_2[/tex]
So, 5 L volume of [tex]SO_2[/tex] present in [tex]\frac{5}{22.4}=0.223[/tex] mole of [tex]SO_2[/tex]
And,
As, 22.4 L volume of [tex]O_2[/tex] present in 1 mole of [tex]O_2[/tex]
So, 3 L volume of [tex]O_2[/tex] present in [tex]\frac{3}{22.4}=0.134[/tex] mole of [tex]O_2[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2SO_2+O_2\rightarrow 2SO_3[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]SO_2[/tex] react with 1 mole of [tex]O_2[/tex]
So, 0.223 moles of [tex]SO_2[/tex] react with [tex]\frac{0.223}{2}=0.1115[/tex] moles of [tex]O_2[/tex]
From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]SO_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]SO_3[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]SO_2[/tex] react to give 2 mole of [tex]SO_3[/tex]
So, 0.223 moles of [tex]SO_2[/tex] react to give 0.223 moles of [tex]SO_3[/tex]
Now we have to calculate the volume of [tex]SO_3[/tex].
As, 1 mole of [tex]SO_3[/tex] gas occupy 22.4 L volume of [tex]SO_3[/tex]
So, 0.223 mole of [tex]SO_3[/tex] gas occupy [tex]0.223\times 22.4=4.99L[/tex] volume of [tex]SO_3[/tex]
Therefore, the volume of the mixture is 4.99 liters.