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2 years ago

What is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.22 kg of h2 and 31.5 kg of n2?

Chemistry

2 answers:

e-lub [12.9K]2 years ago

8 0

Answer : The theoretical yield of ammonia is, 29.58 Kg

Solution : Given,

Mass of $H_2$ = 5.22 Kg = 5220 g

Mass of $N_2$ = 31.5 Kg = 31500 g

Molar mass of $H_2$ = 2 g/mole

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $H_2$ and $N_2$ .

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

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

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

1 mole of $N_2$ react with 3 moles of $H_2$

1125 moles of $N_2$ react with $3\times 1125=3375$ moles of $H_2$

That means $H_2$ is a limiting reagent and $N_2$ is an excess reagent.

Now we have to calculate the moles of ammonia.

From the reaction we conclude that,

3 moles of $H_2$ react to give 2 moles of ammonia

2610 moles of $H_2$ react to give $\frac{2}{3}\times 2610=1740$ moles of ammonia

Now we have to calculate the mass of ammonia.

$\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3$

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

Therefore, the theoretical yield of ammonia is, 29.58 Kg

lilavasa [31]2 years ago

6 0

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

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