Question

n the first step of the industrial process for making nitric acid, ammonia reacts with oxygen in the presence of a suitable catalyst to form nitric oxide and water vapor:4 NH31g2+5 O21g2¡4 NO1g2+6 H2O1g2How many liters of NH31g2 at 850 °C and 5.00 atm are required to react with 1.00 mol of O21g2 in this reaction?

Answer 1

Answer:

10.54 L

Explanation:

Given that:-

Moles of $O_2$ = 1.00 moles

According to the given reaction:-

$4NH_3+7O_2\rightarrow 4NO_2+6H_2O$

7 moles of oxygen gas reacts with 4 moles of ammonia

1 mole of oxygen gas reacts with $\frac{4}{7}$ moles of ammonia

Moles of ammonia = 0.5714 moles

Also, Given that:

Temperature = 850 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (850 + 273.15) K = 1123.15 K  

n = 0.5714 moles

P = 5.00 atm

Using ideal gas equation as:

$PV=nRT$

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

5.00 atm × V = 0.5714 moles ×0.0821 L atm/ K mol  × 1123.15  K  

⇒V = 10.54 L

10.54 L of $NH_3$ required.