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1 month 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:

lions [2.9K]1 month ago

8 0

The calculated theoretical yield of ammonia is 29.58 Kg

In the solution provided,

the mass of $H_2$ is 5.22 Kg, which equals 5220 g

the mass of $N_2$ is 31.5 Kg, equivalent to 31500 g

The molar mass of $H_2$ is 2 g/mole

The molar mass of $N_2$ is 28 g/mole

The molar mass of $NH_3$ is 17 g/mole

First, we must 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 $N_2$ = $\frac{\text{ given mass of }N_2}{\text{ molar mass of }N_2}= \frac{31500g}{28g/mole}=1125moles$

Next, we need to find which reagent is limiting and which is excess.

The balanced equation for the reaction will be,

$N_2+3H_2\rightarrow 2NH_3$

Based on the balanced equation, we find that

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

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

This indicates that $H_2$ is the limiting reagent and $N_2$ is in excess.

Now, we proceed to calculate the moles of ammonia.

According to the reaction, we determine that,

3 moles of $H_2$ will yield 2 moles of ammonia

2610 moles of $H_2$ will provide $\frac{2}{3}\times 2610=1740$ moles of ammonia

Finally, we will calculate the mass of the ammonia produced.

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

$\text{Mass of }NH_3=(1740moles)\times (17g/mole)=29580g=\frac{29580}{1000}=29.58Kg$ Therefore, the theoretical yield of ammonia is 29.58 Kg

VMariaS [2.9K]1 month ago

6 0

The following data is provided:

The mass of H2 is 5.22 kg, which is equal to 5220 g

The mass of N2 is 31.5 kg, or 31500 g

We need to discover:

Theoretical yield of NH3

The explanation is as follows:

The balanced chemical equation is:

N2 + 3H2 → 2NH3

1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3

Calculating moles of N2, we get: # moles of N2 = 31500 g/ 28 g.mol-1 = 1125 moles

Calculating moles of H2, we find: # moles of H2 = 5200 g / 1 g.mol-1 = 5200 moles

Therefore, N2 acts as the limiting reagent

According to stoichiometry:

1 mole of N2 creates 2 moles of NH3

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

Calculating the mass of NH3: Mass of NH3 = 2250 moles * 17 g/mole = 38250 g = 38.3 kg

Thus, the theoretical yield of NH3 equals 38.3 kg

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