Respuesta :
Answer : The maximum volume of [tex]N_2O[/tex] at STP produced is, 116.48 L
Explanation :
First we have to calculate the limiting and excess reagent.
The given balanced chemical reaction is:
[tex]Ca(NO_3)_2(s)+2NH_4Cl(s)\rightarrow 2N_2O(g)+CaCl_2(s)+4H_2O(g)[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]NH_4Cl[/tex] react with 1 mole of [tex]Ca(NO_3)_2[/tex]
So, 5.20 moles of [tex]NH_4Cl[/tex] react with [tex]\frac{5.20}{2}=2.6[/tex] moles of [tex]Ca(NO_3)_2[/tex]
From this we conclude that, [tex]Ca(NO_3)_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]NH_4Cl[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]N_2O[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]NH_4Cl[/tex] react to give 2 mole of [tex]N_2O[/tex]
So, 5.20 mole of [tex]NH_4Cl[/tex] react to give 5.20 mole of [tex]N_2O[/tex]
Now we have to calculate the volume of [tex]N_2O[/tex]
As we know that, 1 mole of substance occupies 22.4 L volume of gas at STP.
As, 1 mole of [tex]N_2O[/tex] occupies 22.4 L volume of gas
So, 5.20 mole of [tex]N_2O[/tex] occupies [tex]5.20\times 22.4L=116.48L[/tex] volume of gas
Thus, the maximum volume of [tex]N_2O[/tex] at STP produced is, 116.48 L
