Question

50kg of nitrogen and 10 kg of hydrogen are mixed to produce ammonia . Calculate the ammonia formed and identify the limiting reagent in the production of ammonia in this situation .

Explain using simple method

..( Define each step ..according to a beginer ..

like ..first balanced chemical equation..converting to moles and identifying limiting reagent etc ( use simple english )

Answer 1

Let us write the balanced equation

N₂ + 3H₂ → 2NH₃

Now calculate the number of moles

Number of moles of N₂ = 50 kg of N₂ = 50 X 10 ³ g/1 kg x 28g = 17.86 x 10 ² mole

Number of moles of H₂ = 10 kg of N₂ = 10 X 10³ g/ 1 kg x 2 = 4.96X 10³ mol

According to the above equation 1 mole of N₂ reacts with 3 moles H₂.

That is 17.86 x 10 ² mole of N₂ reacts with ------moles of H2

= 3/1 X 17.86 x 10 ² = 5.36 x 10³ moles.

Here we have 4.96X 10³ mol of hydrogen. Hence Hydrogen is the limiting reagent.

Let us calculate the amount ammonia formed by reacting 4.96X10³ moles Hydrogen

3 moles of hydrogen -------2 moles of NH₃

4.96 x10³ moles Hydrogen -----?

= 4.96 x10³ X ⅔

= 3.30 x 10³ moles of NH₃