I need help on Chemistry

Question

In the important industrial process for producing ammonia (the Harber Process), the overall reaction is:

N2(g) + 3H2(g) --> 2NH_{3(g) + 100.4 kJ}

A yield of NH₃ of approximately 98% can be obtained at 200 degrees Celsius and 1,000 atmospheres of pressure.

How many grams of N₂ must react to form 1.7 grams of ammonia, NH₃?

Answer 1

N₂(g) + 3H₂(g) --> 2NH₃(g)

1.7 g NH₃ x 1 mole NH₃/17 g = 0.1 moles NH₃ @ 100% yield

Since the yield is only 98%, one would need to increase this to 0.1/0.98 = 0.102 moles NH₃

0.102 moles NH₃ x 1 mole N₂/2 moles NH₃ = 0.0510 moles N₂ needed

0.0510 moles N₂ x 28 g/mole = 1.4 g N₂ needed (to 2 significant figures

(NOTE: Even if you don't correct for the 98% recovery, you will still obtain the correct answer given that you need only 2 significant figures)