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>The elements on the Periodic Table are arranged in order of increasing "Atomic number"
In short, Your Answer would be Option B
Hope this helps!</span>
<h3>
Answer:</h3>
#1. Balanced equation: 2C₅H₅ + Fe → Fe(C₅H₅)₂
#2. Type of reaction: Synthesis reaction
<h3>
Explanation:</h3>
- Balanced equations are equations that obey the law of conservation of mass.
- When an equation is balanced the number of atoms of each element is equal on both side of the equation.
- Equations are balanced by putting appropriate coefficients on the reactants and products.
- In our case, we are going to put coefficients 2, 1 and 1.
- Thus, the balanced equation will be;
2C₅H₅ + Fe → Fe(C₅H₅)₂
- This type of a reaction is known as synthesis reaction, in which two or more reactants or compounds combine to form a single compound or product.
Hey there!
I believe the answer is Combination (or Synthesis) Reaction.
