1 kg of Nitrogen is mixed with 3.5 m3 of hydrogen at 300 K and 101.3 kPa and sent to ammonia converter. The product leaving the converted analyzed 13.7% ammonia, 70.32% hydrogen and 15.98% nitrogen.

(a) Identify the limiting reactant?
(b) What is the percent excess of the excess reactant?
(c) What is the percent conversion of the limiting reactant

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Students

To solve this problem, we need to determine the limiting reactant, the percent excess of the excess reactant, and the percent conversion of the limiting reactant.

(a) To identify the limiting reactant, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced chemical equation. The balanced chemical equation for the reaction is:

N2 + 3H2 -> 2NH3

First, let's calculate the number of moles of each reactant:
1 kg of Nitrogen = 1000 g = 1000/28 = 35.71 moles of Nitrogen

To calculate the moles of hydrogen, we need to use the ideal gas law equation:

PV = nRT

Rearranging the equation to solve for n (number of moles):
n = PV/RT

Given:
P = 101.3 kPa
V = 3.5 m^3
T = 300 K
R = 8.314 J/(mol∙K)

n = (101.3 * 3.5) / (8.314 * 300) = 0.15 moles of Hydrogen

Next, we compare the moles of each reactant to the stoichiometric ratio. The stoichiometric ratio of N2 to H2 is 1:3, which means 1 mole of N2 reacts with 3 moles of H2.

From the stoichiometry, we can see that for every mole of N2, we need 3 moles of H2. However, we have only 0.15 moles of H2 for 35.71 moles of N2. Therefore, Hydrogen is the limiting reactant.

(b) To find the percent excess of the excess reactant, we need to determine the moles of the excess reactant left after the reaction. In this case, Nitrogen is the excess reactant.

We have 35.71 moles of Nitrogen initially. Since 2 moles of NH3 are produced from 1 mole of N2, the maximum number of moles of NH3 that can be produced is 2 * 35.71 = 71.42 moles.

The actual amount of NH3 produced is given as 13.7% of the mixture. Hence, the actual number of moles of NH3 produced is 0.137 * 71.42 = 9.78 moles.

Since 1 mole of NH3 requires 3 moles of H2, the required moles of H2 for the production of 9.78 moles NH3 would be 3 * 9.78 = 29.34 moles.

Therefore, the excess moles of H2 would be 0.15 - 29.34 = -29.19 moles. Since the value is negative, it indicates that there is no excess H2.

So, the percent excess of the excess reactant is 0%.

(c) To find the percent conversion of the limiting reactant (ammonia), we need to compare the actual moles of NH3 produced to the maximum moles of NH3 that could be produced.

From part (b), we found that the actual number of moles of NH3 produced is 9.78 moles. The maximum moles of NH3 that could be produced is 71.42 moles, as calculated earlier.

Therefore, the percent conversion of the limiting reactant (NH3) is (9.78 / 71.42) * 100% = 13.7%.

Hydrogen