Question

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

N2(g) + 3H2(g) → 2NH3(g) + 24,000 calories

A yield of ammonia, NH3, of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure.

How many grams of N2 must react to form 1.7 grams of ammonia?
0.52 g
0.71 g
1.4 g
2.8 g

Answer 1

First the molar mass of ammonia NH3 => 14+3 = 17 g/mole
N2 = 28 grams/mole.

now convert grams of ammonia mole(s)

\[1.7~grams~NH_{3}*(\frac{ mole~NH_{3} }{ 17~grams })~ \frac{ 1.7 }{ 17 }~moles = \frac{ 1 }{ 10 }~moles~NH_{3}\]

\[\frac{ 1 }{ 10 }~NH_{3}*(\frac{ N_{2} }{ 2NH_{3} }) =\frac{ 1 }{ 20 }~mole~NH_{3}*(28\frac{ g }{ mole})=?~grams~N_{2}\]

Answer 2

@marissag