the important industrial process for producing ammonia (the Haber Process), the overall reaction is: N2(g) + 3H2(g) → 2NH3(g) + 100.4 kJ A yield of 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, NH3

Answer 1

Is that 1.7 g NH3 at 98% or 1.7 g NH3 @ 98%? I will assume you want 1.7 g NH3 but with 98% yield.
N2 + 3H2 ==> 2NH3
mols NH3 = 1.7g/molar mass NH3 = 1.7/17 = 0.1 mol.
That will require how many mols N2? That's 0.1 mol NH3 x (1 mol N2/2 mol NH3) = 0.1 x 1/2 = 0.05 mol N2.
How many grams is that? That's 0.05 mol N2 x molar mass N2 = 0.05 x 28 = 1.4 g N2. That is assuming a 100% yield. If you have only 98% yield, you will need 1.4/0.98 = ? g N2. Check my work carefully.

Answer 2

cgjfxjrt

Answer 3

ctyhrjrtthc