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:
0.0253 M/s
Explanation:
From the reaction
N₂ + 3H₂ → 2NH₃
The rate of reaction can be written as
Rate = - ![\frac{d[N_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BN_2%5D%7D%7Bdt%7D)
= - ![\frac{1}{3} \frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
= + ![\frac{1}{2} \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
From the above rate equation we can conclude that the rate of reaction of N₂ is equal to one third of the rate of reaction of H₂,
So,
Rate of reaction of molecular nitrogen = 
Upon calculation, we get rate of reaction of molecular nitrogen = 0.0253 M/s
Answer:
I have for Decompostion
Explanation:
A decomposition reaction occurs when one reactant breaks down into two or more products. It can be represented by the general equation: AB → A + B. In this equation, AB represents the reactant that begins the reaction, and A and B represent the products of the reaction
Answer:
The three isomers having the molecular formula 
are drawn in the figure below.
Explanation:
Answer:
0.68
Explanation:
Number of moles is directly propotionalto the volume at standard condition.
