Chemistry, asked by prasadsunil358, 8 months ago

50.0 kg of N2 and 10.0 kg of H2 are mixed to produce NH3.calculate the NH3 fromed.identify the limiting reagent in the production of NH3 in this situation.​

Answers

Answered by ksudeep2006
0

Answer:

I can't understand your question

Answered by ayanajinn6
3

Answer:

56.1 kg of Ammonia is formed. (The answer may be long but I have explained it step by step.)

Explanation:

The equation goes like this:

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

Calculation of moles:

No. of moles = mass/ molecular mass

No. of moles of N2 = 50000/28 =1786.7 = 17.86 x 10^2 moles

No. of moles of H2 = 10000/2 = 5000 = 50 x 10^2 moles

Finding limiting reagent:

From the above equation:

1 mole of N2 needs 2 moles of H2 to complete the reaction

So 17.86 x 10^2 moles needs x moles of hydrogen

cross multiply:

x = 3 x (17.86 x 10^2) moles

= 53.58 x 10^2 moles

But we have only 50 x 10^2 moles. So hydrogen is the limiting reagent.

Shortcut method to find limiting reagent:

The substance which is less amount is the limiting reagent, because it gets consumed first fully.

Finding the no. of moles of NH3 formed:

3 moles of H2 gives 2 moles of NH3

So, 50 x 10^2 moles of H2 will give x moles of NH3

CROSS MULTIPLY:

x*3 = 2 * 50 * 10^2

x = 100 * 10^2 / 3

x = 10 * 10^3 / 3

x = 3.3 * 10^3

Converting moles into mass:

no. of moles = mass / molecular mass

mass= no. of moles x molecular mass

molecular mass of NH3= 14 + 3 =17g

mass = 3.3 x 10^3 x 17

= 56.1 x 10^3 g

= 56.1 kg

( Since prefix of kg is 10^3)

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