In an experiment Teresa's measures 15.5 mL of water she must have used a
1 answer:
You might be interested in
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
Answer:
4.767 grams of KCl are produced from 2.50 g of K and excess Cl2
Explanation:
The balanced equation is
2 K+ Cl2 --->2 KCI
Here the limiting agent is K. Hence, the amount of KCl will be calculated as per the mass of 2.50 gram of K
Mass of one atom/mole of potassium is 39.098 grams
Number of moles is 2.5 grams =
So, 2 moles of K produces 2 moles of KCL
0.064 moles of K will produces 0.064 moles of KCl
Mass of one molecule of KCl is 74.5513 g/mol
Mass of 0.064 moles of KCl is 4.767 grams