Question

solve the problem
step by step​

Answer 1

mass of N 2 = 50 kg = 5 ×10^4 g

molar mass of N 2 = 28 g

moles of N 2

= mass / molar mass

= 5×10^4 / 28

=0.178×10^4

=1.78×10^3

mass of H 2=10 kg = 10×10^3 g =10^4 g

molar mass of H 2 = 2

moles of H 2

= 10^4/2

= 10×10^3/2

=5×10^3

N 2 +.3 H 2 ----> 2 NH 3

From this we can conclude that 1 mole of nitrogen combines with 3 moles of H 2 to give 2 moles of ammonia

So , 1.78×10^4 moles of nitrogen will react with

1.78×10^4×3 =5.34×10^3 g Hydrogen

amt of Hydrogen available = 5 ×10^3

hence , Hydrogen is the limiting reagent

3 moles of Hydrogen react to give 2 moles of ammonia

so , 5×10^3 moles of Hydrogen will react with

5×10^3 × 2 / 3 = 3.33 ×10 ^ 3 moles of ammonia

ANSWER:

Hydrogen Is the limiting reagent

Amount of ammonia formed=3.33×10^3 moles