Question

2 MnO₄ - (aq) + 5 C₂O₄²⁻ (aq) + 16 H⁺ (aq) → 2 Mn²⁺ (aq) + 10 CO₂ (g) + 8 H₂O (l)Permanganate and oxalate ions react in an acidified solution according to the balanced equation above. How many moles of CO₂(g) are produced when 20. mL of acidified 0.20 M KMnO₄ solution is added to 50. mL of 0.10 M Na₂C₂O₄ solution?

Answer 1

Answer:

The produced moles are 1,0×10⁻²CO₂.

Explanation:

The first thing we should know is what reagent is the limiting one. To know this, it is necessary to obtain moles, thus:

0,020 KMnO₄ liters  × $\frac{0,20mol}{L}$ = 4,0×10⁻³ KMnO₄ moles

0,050 Na₂C₂O₄ liters  × $\frac{0,10mol}{L}$ = 5,0×10⁻³  Na₂C₂O₄ moles

The global reaction is:

2 MnO₄⁻ (aq) + 5 C₂O₄²⁻ (aq) + 16 H⁺ (aq) → 2 Mn²⁺ (aq) + 10 CO₂ (g) + 8 H₂O

Thus, two moles of MnO₄⁻ reacts with five moles of C₂O₄²⁻. It means that for a complete reaction of 4,0×10⁻³ KMnO₄ moles you need:

4,0×10⁻³ KMnO₄ moles × $\frac{5 Na2C2O4 moles}{2 KMnO4 moles}$ =

1,0×10⁻²Na₂C₂O₄ moles but there are just 5,0×10⁻³  Na₂C₂O₄ moles. Thus, limiting reagent is Na₂C₂O₄.

Now, the produced CO₂ moles are calculated with the limiting reagent moles, knowing that 5 C₂O₄⁻ moles produce 10 CO₂ moles, thus:

5,0×10⁻³  Na₂C₂O₄ moles × $\frac{10 CO2 mol}{5 C2O4- moles}$ =

1,0×10⁻²CO₂ moles

I hope it helps!