Question

50kg of N2 and 10kg of H2 reacts to give NH3.Identify the limiting reagent and calculate the amount of NH3 produced​

Answer 1

Answer:

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Explanation:

2N2 + 3H2 --------> 2NH3

From the above reaction we get to know that 2 moles of N2 reacts with 3 moles of H2 to produce 2 moles of NH3

Given mass of N2 is 50,000 g

No of moles of N2 = 50,000/28 = 1785.714 moles

Given mass of H2 is 10,000 g

No of moles of H2 = 10,000/2 = 5,000 moles

As N2 is less in number therefore N2 is limiting reagent

For 2 moles of N2 we get 2 moles of NH3

Therefore,

For 1785.714 moles of N2 we get 1785.714 moles of NH3

1785.714 = Mass of NH3/(14 + 3)

1785.714 * 17 = Mass of NH3

Mass of NH3 = 30357.14 g

Answer 2

Answer:

n2 is the limiting reagent