Answer:
%WO₃ = 20.84%
Explanation:
The half-reaction for the reduction of WO₃ is:
a) W⁺⁶ +3e⁻ ⇒ W⁺³
The reaction that takes place in the analysis is
MnO₄⁻ + W⁺³ ⇒ Mn⁺² + WO₂⁺²
In order to balance this reaction, we write the two half-reactions:
W⁺³ ⇒ WO₂⁺²
2H₂O + W⁺³ ⇒ WO₂⁺²
2H₂O + W⁺³ ⇒ WO₂⁺² + 4H⁺
2H₂O + W⁺³ ⇒ WO₂⁺² + 4H⁺ + 3e⁻ b)
MnO₄⁻ ⇒ Mn⁺²
MnO₄⁻ ⇒ Mn⁺² + 4H₂O
MnO₄⁻ + 8H⁺ ⇒ Mn⁺² + 4H₂O
MnO₄⁻ + 8H⁺ + 5e⁻ ⇒ Mn⁺² + 4H₂O c)
To balance charges, we multiply half-reaction b) by 5, and half-reaction c) by 3; and add them:
10H₂O + 5W⁺³ + 3MnO₄⁻ + 24H⁺ + 15e⁻ ⇒ 5WO₂⁺² + 20H⁺ + 15e⁻ + 3Mn⁺² + 12H₂O
We simplify the equation and we're left with the balanced reaction:
5W⁺³ + 3MnO₄⁻ + 4H⁺⇒ 5WO₂⁺² + 3Mn⁺² + 2H₂O d)
To determine the moles of W⁺³ in the aliquot, we make use of reaction d), using the known moles of MnO₄⁻ that reacted with the sample:
[tex]0.01143 L * 0.09713 M *\frac{5molesW}{3molesMnO4} =0.00185 moles[/tex]
Now we determine the moles of W⁺³ in the sample of 500 mL:
0.00185 moles * 500.0 mL / 100.0 mL = 0.00925 moles
Due to half-reaction a), we know that the moles of W⁺³ are equal to the moles of WO₃ in the sample of the vase. Now we determine the mass of WO₃ in the sample, using the molecular weight:
0.00925 moles WO₃ * 231,84 g /mol= 2.145 g WO₃
All that's left now is to calculate the percent of WO₃ in the sample:
[tex]\frac{2.145 g}{10.29g}[/tex] * 100% = 20.84%
