Question

Sodium oxalate, Na2C2O4, in solution is oxidized to CO2(g) by MnO−4 which is reduced to Mn2+. A 50.3 −mL volume of a solution of MnO−4 is required to titrate a 0.342 −g sample of sodium oxalate. This solution of MnO−4 is then used to analyze uranium-containing samples. A 4.62 −g sample of a uranium-containing material requires 32.5 mL of the solution for titration. The oxidation of the uranium can be represented by the change UO2+→UO2+2.

Calculate the percentage of uranium in the sample.

Answer 1

Answer:

the percentage of uranium in the sample is 8.4957%

Explanation:

the solution is attached in the Word file

Answer 2

Oxidation is the process of the addition of oxygen to an element or molecule. The percentage of uranium in the sample is 8.4957%.

What is oxidation-reduction?

In oxidation, the electrons are lost by the atom or the molecule whereas in reduction electrons are gained by the species.

The chemically balanced reaction is:

$\rm 5C_{2}O_{4}^{2-} + 2MnO_{4}^{-} + 16H^{+} \rightarrow 10CO_{2} + 2Mn^{2+} + 8H_{2}O$

From the reaction, it can be seen that 5 moles of $\rm 5C_{2}O_{4}^{2-}$reacts with 2 moles of $\rm 2MnO_{4}^{-}.$

Moles of Sodium oxalate are calculated as:

$\rm n = 0.342 \times (\dfrac{1}{133.99}) = 2.552 \times 10^{-3} \;\rm moles$

Moles of permanganate are calculated as:

$\rm n = 2.552 \times 10^{-3} \times (\dfrac{2}{5}) = 1.021 \times 10^{-3}\;\rm moles$

From moles the molarity of permanganate is calculated as:

$\rm \dfrac{1.021 \times 10^{-3}\;\rm moles}{50.3 \;\rm mL} \times \dfrac{1000\;\rm ml}{1\;\rm L} = 2.03 \times 10^{-2}\;\rm moles /L$

The balanced reaction between permanganate and uranyl ion is shown as:

$\rm 5UO^{2+} + 2MnO_{4}^{-} + 6H^{+}\rightarrow 5UO_{2}^{+2} + 2Mn^{2+} + 3H_{2}O$

From the above reaction moles of uranyl ion are calculated as:

$32.5 \times \dfrac{1}{1000} \times \dfrac{2.03 \times 10^{-2}}{1}(\dfrac{5}{2}) = 1.649 \times 10^{-3} \;\rm moles$

The mass of uranium is calculated as:

$1.649 \times 10^{-3} \;\rm moles \times \dfrac {238\;\rm g}{1\;\rm mol} = 0.3925 \;\rm g$

The percent is calculated as:

$\dfrac{0.3925}{4.62}\times 100 = 8.49 \%$

Therefore, 8.49% is the percentage of uranium in the sample.

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