We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, 
moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.
<span>1) 0.2M ferric nitrate is added gradually to 1M sodium hydroxide. In result, a red precipitate appears. The precipitate is ferric hydroxide.
2) </span><span>0.2M potassium chromate is added gradually to 0.05M lead acetate. in result, a yellow precipitate appears. The precipitate is called potassium acetate.
The common between the two is that the colors originated from one of the reactants. </span>
The answer is the 3rd choice
6.4mole•64.06g/1mole=409.98g
DeltaH formation = deltaH of broken bonds - deltaH of formed bonds
Broken bonds: tiple bond N-N and H-H bond
Formed bonds: N-H and N-N bonds
You also have to take note of the molar coefficients
deltaH formation = <span> [(N≡N) + 2 * (H-H)] - [4 * (N-H) + (N-N)]
= (945 + 2*436) - (4*390 + 240)
= 17 kJ/mol
The answer is 17 kJ/mol.</span>
