The reaction will be
Ag2SO4 + 2KCl --> 2AgCl (s) + K2SO4
We know that
Molarity = Number of moles / Volume of solution in litres
Molarity of KCl = 0.5
Volume = 25 mL
so moles of KCl = molarity X volume = 0.5 X 25 mL = 12.5 millimoles
As per stoichiometry of reaction 1 mole of Ag2SO4 will react with two moles of KCl to form 2 moles of silver chloride
so moles of AgCl formed = 2 X 12.5 millimoles = 25 millimoles
The moles of Ag2SO4 reacted = 12.5 / 2 millimoles = 6.25 millimoles
Volume used = 45 mL
so molarity of Ag2SO4 = moles / volume = 6.25 / 45 = 0.139 molar (original concentration of silver sulphate)
The sulphate remained will be = moles of Ag2SO4 used = 0.139 molar
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