Question

plz answer the above question

Answer 1

et 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 3g/1 kg x 28g = 17.86 x 10 2 mole

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

According to the above equation 1 mole of N2reacts 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

Answer 2

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Moles of N2 = mass/ molar mass

Moles of N2 = 56 / 28

Moles of N2 = 2mol

Moles of H2 = mass/molar mass

Moles of H2 = 10 /2

Moles of H2 = 5 mol

Now balanced Equation is :-

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


It means that 1 mole N2 forms 2 mole NH3

Then .......

2 mole N2 will form 4 moles of NH3

So 4 moles of NH3 has mass :-

Moles of NH3 = mass / molar mass

Molar mass of NH3 = 14 + 1×3 = 17 g/ mol

Mass of NH3 = moles of NH3 × molar mass

Mass of NH3 = 4 × 17

Mass of NH3 = 68 mol

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