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?

Question

Chemistry

Pogo

1 September, 01:26

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?

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Answers (1)

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Hugo Armstrong · Answer 1

There will be produced 0.01 moles of CO2

Explanation:

Step 1: Data given

volume of 0.20 M KMnO4 solution = 20 mL = 0.02 L

Volume of 0.10 M Na2C2O4 solution = 50 mL = 0.05 L

Step 2: The balanced equation

2 MnO4 - (aq) + 5C2O4^2 - (aq) + 16 H + (aq) → 2 Mn^2 + (aq) + 10CO2 (g) + 8H2O (l)

Step 3: Calculate moles of KMnO4

moles KMnO4 = Molarity * volume

moles KMnO4 = 0.20 M * 0.02 L

moles KmnO4 = 0.004 moles

Step 4: Calculate moles of Na2C2O4

Moles Na2C2O4 = 0.10 M * 0.05 L

Moles Na2C2O4 = 0.005 moles

Step 5: Calculate limiting reactant

For 2 moles KMnO4 we need 5 moles Na2C2O4

Na2C2O4 is the limiting reactant

There will react 0.005 moles

KMnO4 is in excess. There will react 0.005 / 2.5 = 0.002 moles

Step 6: Calculate moles of CO2

For 2 moles KMnO4 we need 5 moles Na2C2O4, there will be produced 10 moles of CO2

For 0.005 moles of Na2C2O4 there will be produced 0.01 moles of CO2

There will be produced 0.01 moles of CO2