Answer: The correct answer is option A.
Explanation:
100 mL of 1.0 M
Volume of = 100 mL = 0.1 L
Moles of = n
1 mole of gives 3 moles of sodium ions and 1 mole of phosphate ions.
Moles of sodium ions =
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of sodium ions=
100 mL of 1.0 M
Volume of = 100 mL = 0.1 L
Moles of = n'
1 mole of gives 1 mole of silver ions and 1 mole of nitrate ions.
According to reaction 1 mole of reacts with 3 moles of
.
Then 0.1 mole of will react with:
of
Moles of sodium phosphate left unreacted = 0.1 mol - 0.0333 mol = 0.0667 mol
1 mole of gives 3 moles of sodium ions and 1 mole of phosphate ions.
As we can see that silver nitrate is in limiting amount
According to reaction 3 mole of gives with 1 mole of
.
So, when 0.1 mol of reacts it gives:
of
Moles of phosphate ions left in solution=
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of phosphate ions=
Moles of nitrate ions =
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of nitrate ions=
But is an excessive reagent its concentration will be less
