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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
Answer:
0.40 L
Explanation:
Calculation of the moles of 
as:-
Mass = 51.24 g
Molar mass of = 171.34 g/mol
The formula for the calculation of moles is shown below:
Thus,
Volume = 1.20 L
The expression for the molarity is:
Thus,
Considering
Given that:
So,
<u>The volume of 0.24925M stock solution added = 0.40 L
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