Question

how many ml of 0.032 molar kmno4 are required to react with 50.0 ml of 0.100 molar h2c2o4 in the presence of excess h2so4

Answer 1

62.5 ml of 0.032 M KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄

Step-by-step explanation:

The balanced reaction is:

2 KMnO₄ + 5 H₂C₂O₄ + 3 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 8 H₂O + 10 CO₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• KMnO₄: 2 moles
• H₂C₂O₄: 5 moles
• H₂SO₄: 3 moles
• K₂SO₄: 1 mole
• MnSO₄: 2 moles
• H₂O: 8 moles
• CO₂: 10 moles

Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

`Molarity=(number of moles of solute)/(volume)`

Molarity is expressed in units
`(moles)/(liter)`

In this case, 50 mL (0.05 L) of 0.1 M H₂C₂O₄ react. So, replacing the data in the definition of molarity:

`0.1 M=(number of moles of solute)/(0.05 L)`

Solving:

number of moles of solute= 0.1 M*0.05 L

number of moles of solute= 0.005 moles

So, 0.005 moles of H₂C₂O₄ react. Then you can apply the following rule of three: if by stoichiometry 5 moles of H₂C₂O₄ react with 2 moles of KMnO₄, 0.005 moles of H₂C₂O₄ react with how many moles of KMnO₄?

`moles of KMnO_(4) =(0.005moles of H_(2) C_(2) O_(4)* 2moles of KMnO_(4) )/(5moles of H_(2) C_(2) O_(4) )`

moles of KMnO₄= 0.002 moles

Knowing that the molarity of KMnO₄ is 0.032 M, replacing in its definition and solving:

`0.032 M=(0.002 moles)/(volume)`

`volume=(0.002 moles)/(0.032 M)`

volume= 0.0625 L= 62.5 mL

62.5 ml of 0.032 M KMnO₄ are required to react with 50.0 ml of 0.100 molar H₂C₂O₄ in the presence of excess H₂SO₄