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

The produced moles are 1,0×10⁻²CO₂.

Step-by-step 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 ×
`(0,20mol)/(L)` = 4,0×10⁻³ KMnO₄ moles

0,050 Na₂C₂O₄ liters ×
`(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 ×
`(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 ×
`(10 CO2 mol)/(5 C2O4- moles)` =

1,0×10⁻²CO₂ moles

I hope it helps!