Answer :
The concentration of ClO₃⁻ in the resulting solution is 0.447 M
How to determine the mole of ClO₃⁻ in AgClO₃
We'll begin by obtaining the mole of AgClO₃ in the solution. This is illustrated as follow:
- Volume = 821 mL = 821 / 1000 = 0.821 L
- Molarity of AgClO₃ = 0.349 M
- Mole of AgClO₃ =?
Molarity = mole / volume
Cross multiply
Mole = molarity × volume
Mole of AgClO₃ = 0.349 × 0.821
Mole of AgClO₃ = 0.287 mole
Finally, we shall determine the mole of ClO₃⁻ in AgClO₃ solution as folllow:
AgClO₃(aq) <=> Ag⁺(aq) + ClO₃⁻(aq)
From the balanced equation above,
1 mole of AgClO₃ contains 1 mole of ClO₃⁻
Therefore,
0.287 mole of AgClO₃ will also contain 0.287 mole of ClO₃⁻
How to determine the mole of ClO₃⁻ in Mn(ClO₃)₂
We'll begin by obtaining the mole of Mn(ClO₃)₂ in the solution. This is illustrated as follow:
- Volume = 913 mL = 913 / 1000 = 0.913 L
- Molarity of Mn(ClO₃)₂ = 0.267 M
- Mole of Mn(ClO₃)₂ =?
Molarity = mole / volume
Cross multiply
Mole = molarity × volume
Mole of Mn(ClO₃)₂ = 0.267 × 0.913
Mole of Mn(ClO₃)₂ = 0.244 mole
Finally, we shall determine the mole of ClO₃⁻ in Mn(ClO₃)₂ solution as folllow:
Mn(ClO₃)₂(aq) <=> Mn²⁺(aq) + 2ClO₃⁻(aq)
From the balanced equation above,
1 mole of Mn(ClO₃)₂ contains 2 mole of ClO₃⁻
Therefore,
0.244 mole of Mn(ClO₃)₂ will contain = 0.244 × 2 = 0.488 mole of ClO₃⁻
How to determine the molarity of ClO₃⁻ in the resulting solution
- Mole of ClO₃⁻ in AgClO₃ = 0.287 mole
- Mole of ClO₃⁻ in Mn(ClO₃)₂ = 0.488 mole
- Total mole of ClO₃⁻ = 0.287 + 0.488 = 0.775 mole
- Total volume = 0.821 + 0.913 = 1.734 L
- Molarity of ClO₃⁻ =?
Molarity = mole / volume
Molarity of ClO₃⁻ = 0.775 / 1.734
Molarity of ClO₃⁻ = 0.447 M
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