You can calculate the excess reactant by subtracting the mass of excess reagent consumed from the total mass of reagent given therefore,
The answer: Theoretical yield is 121.60 g of NH₃
Excess reactant is H₂
Rate limiting reactant is N₂
explanation: 100 g of Nitrogen
100 g of hydrogen
We are required to identify the theoretical yield of the reaction, the excess reactant and the rate limiting reagent.
We first write the equation for the reaction between nitrogen and hydrogen;
N₂ + 3H₂ → 2NH₃
From the reaction 1 mole of nitrogen reacts with 3 moles of Hydrogen gas.
Secondly we determine the moles of nitrogen gas given and hydrogen gas given;
Moles of Nitrogen gas
Moles = Mass ÷ Molar mass
Molar mass of nitrogen gas = 28.0 g/mol
Moles of Nitrogen gas = 100 g ÷ 28 g/mol 3.57 moles
Moles of Hydrogen gas
Molar mass of Hydrogen gas = 2.02 g/mol
Moles = 100 g ÷ 2.02 g/mol
= 49.50 moles
From the mole ratio given by the equation, 1 mole of nitrogen requires 3 moles of Hydrogen gas.
Thus, 3.57 moles of Nitrogen gas requires (3.57 × 3) 10.71 moles of Hydrogen gas.
This means, Nitrogen gas is the rate limiting reagent and hydrogen gas is the excess reactant.
Third calculate the theoretical yield of the reaction.
1 mole of nitrogen reacts to from 2 moles of ammonia gas
Therefore;
Moles of ammonia gas produced = Moles of nitrogen × 2
= 3.57 moles × 2
= 7.14 moles
But; molar mass of Ammonia gas is = 17.03 g/mol
Therefore;
Mass of ammonia gas produced = 7.14 moles × 17.03 g/mol
= 121.59 g
= 121.60 g
Thus, the theoretical amount of ammonia gas produced is 121.60 g
Nothing, he shouldn’t be able to move it. Think about it like this say you try really hard to push something that is 5,000 pounds and you push as hard as you can. Well you can’t move it bc it weighs more than you can push. I’m sure their is a equation you can use to see how much you can push (body weight=force?)
Answer:
No.
Explanation:
The reason comes the <em>Law of Conservation of Mass</em>.
In an ordinary chemical reaction, <em>you cannot create or destroy atoms</em>.
So, you must have as many atoms at the beginning of a reaction (in the reactants) as at the end (in the products)
We use this principle to balance chemical equations.
For example, the equation for the formation of water from hydrogen and oxygen is
2H₂ + O₂ ⟶ 2H₂O
There are four atoms of H and two of O both before and after the reaction.
3-ethyl-5,6-dimethyloctane
3-ethyl-2,3-dimethylpentane
