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.
Answer:
i think the answer is c hoped this helped
Explanation:
We can skip option B and D because NaCl is salt and H₂SO₄ is a strong acid.
Neutralization reactions are those reactions in which acid and base react to form salt and water.
As water being amphoteric in nature can react with HCl as follow,
HCl + H₂O ⇆ H₃O⁺ + OH⁻
In this case no salt is formed, so we can skip this option.
Ammonia being a weak base can abstract proton from HCl as follow,
HCl + NH₃ → NH₄Cl
Ammonium Chloride is a salt. So, among all four options, Option-C is the correct answer.
<span>when the number of moles Ca = mass of Ca / molar mass of Ca.
and we can get the molar mass of Ca, it is = 40 g/mol
and we have already the mass of Ca (given) = 9.8 g
so, by substitution: the moles Ca = 9.8 g / 40 g/mol
= 0.245 moles</span>
