Respuesta :

Given:

Mass of H2 = 5.22 kg = 5220 g

Mass of N2 = 31.5 kg = 31500 g

To determine:

Theoretical yield of NH3

Explanation:

The balanced chemical reaction is:

N2 + 3H2 → 2NH3

1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3

# moles of N2 = 31500 g/ 28 g.mol-1 = 1125 moles

# moles of H2 = 5200 g/ 1 g.mol-1 = 5200 moles

Therefore N2 is the limiting reagent

Based on the stoichiometry:

1 mole of N2 forms 2 moles of NH3

Thus, 1125 moles of N2 will yield : 1125 * 2 = 2250 moles of NH3

Mass of NH3 = 2250 moles * 17 g/mole = 38250 g = 38.3 kg

Ans: Theoretical yield of NH3 = 38.3 kg

Answer : The  theoretical yield of ammonia is, 29.58 Kg

Solution : Given,

Mass of [tex]H_2[/tex] = 5.22 Kg  = 5220 g

Mass of [tex]N_2[/tex]= 31.5 Kg = 31500 g

Molar mass of [tex]H_2[/tex] = 2 g/mole

Molar mass of [tex]N_2[/tex] = 28 g/mole

Molar  mass of [tex]NH_3[/tex] = 17 g/mole

First we have to calculate the moles of [tex]H_2[/tex] and [tex]N_2[/tex] .

Moles of [tex]H_2[/tex]= [tex]\frac{\text{ given mass of }H_2}{\text{ molar mass of }H_2}= \frac{5220g}{2g/mole}=2610moles[/tex]

Moles of [tex]N_2[/tex] = [tex]\frac{\text{ given mass of }N_2}{\text{ molar mass of }N_2}= \frac{31500g}{28g/mole}=1125moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

From the balanced reaction we conclude that

1 mole of [tex]N_2[/tex] react with 3 moles of [tex]H_2[/tex]

1125 moles of [tex]N_2[/tex] react with [tex]3\times 1125=3375[/tex] moles of [tex]H_2[/tex]

That means [tex]H_2[/tex] is a limiting reagent and [tex]N_2[/tex] is an excess reagent.

Now we have to calculate the moles of ammonia.

From the reaction we conclude that,

3 moles of [tex]H_2[/tex] react to give 2 moles of ammonia

2610 moles of [tex]H_2[/tex] react to give [tex]\frac{2}{3}\times 2610=1740[/tex] moles of ammonia

Now we have to calculate the mass of ammonia.

[tex]\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3[/tex]

[tex]\text{Mass of }NH_3=(1740moles)\times (17g/mole)=29580g=\frac{29580}{1000}=29.58Kg[/tex]

Therefore, the  theoretical yield of ammonia is, 29.58 Kg