Answer:
2Ag(s) + 1H2S(g) → 1Ag2S(s) + 1H2(g)
Explanation:
Step 1: Data given
Ag(s) + H2S(g) → Ag,S(s) + H2(g)
Ag has and oxidation number of +1
S has an oxidation number of -2
H has an oxidation number of +1
Step 2: The balanced equation
Ag+ + 2H+ + S^2- →
The silver ion and the sulfur ion will bind, but we need 2 silver atoms to bind with 1 sulfur atom.
2 H- atoms will appear as H2.
The balanced equation is:
2Ag+ + 2H+ +S^2- → Ag2S + H2
2Ag(s) + 1H2S(g) → 1Ag2S(s) + 1H2(g)
yes 89.75 sorry I'm not the best at math you should look on the internet
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.