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

Answer:

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

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:

$Molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$

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

$0.1 M=\frac{number of moles of solute}{0.05 L}$

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?

$moles of KMnO_{4} =\frac{0.005moles of H_{2} C_{2} O_{4}* 2moles of KMnO_{4} }{5moles of H_{2} C_{2} O_{4} }$

moles of KMnO₄= 0.002 moles

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

$0.032 M=\frac{0.002 moles}{volume}$

$volume=\frac{0.002 moles}{0.032 M}$

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