Question

Ammonia and oxygen produce nitrogen dioxide and water. What volume of nitrogen dioxide gas will be produced if $30.5$ grams of ammonia is reacted with excess oxygen?

Answer 1

Hint: Ammonia is an inorganic compound which has a very characteristic pungent smell. Ammonia is highly water soluble and toxic to the human body. Ammonia is highly reactive in nature and combines with oxygen in different proportions to form different oxides of nitrogen.

Complete Step By Step Answer:
 When ammonia reacts with excess oxygen it forms nitrogen dioxide and water. this chemical reaction is expressed in balanced state as:
 $4N{H}_3+7O_{}\to 4N{O}_2+6H_{2}O$
From the above chemical reaction we see that for every $4$ moles of $N{H}_3$ , $4$ moles of $N{O}_2$ is produced. Hence the ratio of both the compounds is $1:1$ .
Mass of ammonia is already given in question which is $30.5$ grams.
Then to calculate the total moles of ammonia in a given mass, we have to divide the mass of ammonia with its molecular mass.
Molecular mass of ammonia $N{H}_3$ $=$ $1\times$ mass of nitrogen $+$ $3\times$ mass of hydrogen
Molecular mass of ammonia $N{H}_3$ $=$ $14+3$
After solving the above equation, we get
Molecular mass of ammonia $N{H}_3$ $=$ $17$ grams
Moles of ammonia $={\displaystyle \frac{m}{M}}$
Where $m=$ mass of ammonia
 $M=$ Molecular mass of ammonia
Put the values in formula of mole
Moles of ammonia $={\displaystyle \frac{30.5}{17}}$
After solving the above equation, we get
Moles of ammonia $=1.794$ moles
From the concept of molar volume, we know that the volume of one mole of ammonia at NTP will be equal to $22.4$ litres.
The volume of $1$ mole $N{H}_3$ $=$ $22.4$ litres of ammonia
But we have a total of $1.794$ moles of ammonia. So, volume of ammonia will become
Volume of $1.79$ moles of ammonia $=22.4\times 1.794$ litres of ammonia
After calculation, we find that
Volume of $1.794$ moles of ammonia $=40.18$ litres of ammonia
From the chemical equation we see that the ratio between ammonia and nitrogen dioxide is $1:1$ then the number of moles of ammonia used and number of moles of nitrogen dioxide produced is the same.
Moles of $N{H}_3$ $=$ moles of $N{O}_2$
Moles of $N{O}_2$ $=$ $1.794$ moles
Volume of $1$ mole of $N{O}_2$ at NTP is $22.4$ litres, then volume of $1.794$ moles of nitrogen dioxide will be
Volume of $N{O}_2$ $=22.4\times 1.794$ litres of nitrogen dioxide
After solving, we get
Volume of $1.794$ moles of nitrogen dioxide $=40.18$ litres of nitrogen dioxide
 $\Rightarrow$ Hence, $40.18$ litres volume of nitrogen dioxide gas will be produced if $30.5$ grams of ammonia is reacted with excess oxygen.

Note:
Volume of particles of any element equal to Avogadro’s number occupies $22.4$ litres of volume under the NTP condition. Alternative method to calculate volume of gas is by using the ideal gas equation which is expressed mathematically as $PV=nRT$ .