Calcium nitrate will react with ammonium chloride at slightly elevated temperatures, as represented in the equation below. Ca(NO3)2(s) + 2NH4Cl(s) → 2N2O(g) + CaCl2(s) + 4H2O(g) What is the maximum volume of N2O at STP that could be produced using a 5.20-mol sample of each reactant?

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26-02-2020
Chemistry

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Calcium nitrate will react with ammonium chloride at slightly elevated temperatures, as represented in the equation below. Ca(NO3)2(s) + 2NH4Cl(s) → 2N2O(g) + CaCl2(s) + 4H2O(g) What is the maximum volume of N2O at STP that could be produced using a 5.20-mol sample of each reactant?

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Alleei Alleei · Answer 1 · 2020-02-27T07:00:14+01:00

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