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
Independent and dependent variables and constants
Answer is: 5,75·10⁻¹.
Kf = 2,3·10⁶ 1/s.
K = 4,0·10⁸ 1/s.
Kr = ?
Kf - <span>forward rate constant.
K - </span><span>equilibrium constant.
Kr - </span><span>reverse rate constant.
</span>Since both Kf and Kr are constants at a given temperature, their ratio is also a constant that
is equal to the equilibrium constant K.<span>
K = Kf/Kr.
Kr = Kf/K = </span>2,3·10⁶ 1/s ÷ 4,0·10⁸ 1/s = 5,75·10⁻¹.
Answer:
The coefficients are 2 for H₂O and 1 for Ca(OH)₂.
Explanation:
Let's consider the following reaction.
Ca(OH)₂(aq) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂O(l)
According to the balanced equation, the molar ratio of H₂O to Ca(OH)₂ is 2:1. Using this conversion factor, we have the following proportion:
moles Ca(OH)₂. (2 mol H₂O ÷ 1 mol Ca(OH)₂) = moles H₂O
