Question

ammonium nitrate decomposes as follow: NH4NO3 ==> N20 + 2H2O Calculate : a. the volume of nitrous oxide formed when 40 gram of ammonium nitrate is heated. b. Mass of water in grams when 54 grams of ammonium nitrate is heated ​

Answer 1

(a) The volume of nitrous oxide formed is, 11.2 L

(b) The mass of water is, 24.3 grams.

Explanation :

The given balanced chemical reaction is:

$NH_4NO_3\rightarrow N_2O+2H_2O$

Part a :

First we have to calculate the moles of $NH_4NO_3$

$\text{Moles of }NH_4NO_3=\frac{\text{Mass of }NH_4NO_3}{\text{Molar mass of }NH_4NO_3}$

Molar mass of $NH_4NO_3$ = 80 g/mole

$\text{Moles of }NH_4NO_3=\frac{40g}{80g/mole}=0.5mole$

Now we have to calculate the moles of $N_2O$

From the balanced chemical reaction we conclude that,

As, 1 mole of $NH_4NO_3$ decompose to give 1 mole of $N_2O$

So, 0.5 mole of $NH_4NO_3$ decompose to give 0.5 mole of $N_2O$

Now we have to calculate the volume of $N_2O$ formed.

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

As, 1 mole of $N_2O$ occupies 22.4 L volume

So, 0.5 mole of $N_2O$ occupies 0.5 × 22.4 L = 11.2 L volume

Thus, the volume of nitrous oxide formed is, 11.2 L

Part b :

First we have to calculate the moles of $NH_4NO_3$

$\text{Moles of }NH_4NO_3=\frac{\text{Mass of }NH_4NO_3}{\text{Molar mass of }NH_4NO_3}$

Molar mass of $NH_4NO_3$ = 80 g/mole

$\text{Moles of }NH_4NO_3=\frac{54g}{80g/mole}=0.675mole$

Now we have to calculate the moles of $H_2O$

From the balanced chemical reaction we conclude that,

As, 1 mole of $NH_4NO_3$ decompose to give 2 mole of $H_2O$

So, 0.675 mole of $NH_4NO_3$ decompose to give 2 × 0.675 = 1.35 mole of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=1.35mol\times 18g/mol=24.3g$

Thus, the mass of water is, 24.3 grams.

Learn more about : Stoichiometry

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