Question

Hey frnds plssssss help me ....... plsssssss plsssssss it's very urgent plssss :-(

Answer 1

Dihydrogen is limiting reagent
3.30×10^3 mol NH3 is obtained

Answer 2

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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