Am letting the picture doing the talk.
H₂SO₄:
V=0,95L
Cm=0,420mol/L
n = CmV = 0,42mol/L * 0,95L = 0,399mol
KOH:
V=0,9L
Cm=0,26mol/L
n = CmV = 0,26mol/L * 0,9L = 0,234mol
H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O
1mol : 2mol
0,399mol : 0,234mol
limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol
n = 0,282mol
V = 0,950L + 0,900L = 1,85L
Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
22.7 liters
The molar volume of an ideal gas depends on the temperature and pressure. One mole of any ideal gas occupies 22.7 liters at 0 0C and 1 bar (STP).
Hope this helped
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb