Question

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?

Answer 1

Answer : The maximum volume of
`N_2O` at STP produced is, 116.48 L

Explanation :

First we have to calculate the limiting and excess reagent.

The given balanced chemical reaction is:

`Ca(NO_3)_2(s)+2NH_4Cl(s)\rightarrow 2N_2O(g)+CaCl_2(s)+4H_2O(g)`

From the balanced reaction we conclude that

As, 2 mole of
`NH_4Cl` react with 1 mole of
`Ca(NO_3)_2`

So, 5.20 moles of
`NH_4Cl` react with
`(5.20)/(2)=2.6` moles of
`Ca(NO_3)_2`

From this we conclude that,
`Ca(NO_3)_2` is an excess reagent because the given moles are greater than the required moles and
`NH_4Cl` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`N_2O`

From the reaction, we conclude that

As, 2 mole of
`NH_4Cl` react to give 2 mole of
`N_2O`

So, 5.20 mole of
`NH_4Cl` react to give 5.20 mole of
`N_2O`

Now we have to calculate the volume of
`N_2O`

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

As, 1 mole of
`N_2O` occupies 22.4 L volume of gas

So, 5.20 mole of
`N_2O` occupies
`5.20* 22.4L=116.48L` volume of gas

Thus, the maximum volume of
`N_2O` at STP produced is, 116.48 L