Question

molecular equation for the re Dante sider the following mo at: (Place a and b at the er It not completed before lab begins) on for the reaction Involved in today's the end of the Procedure for Part 2) MnSO4 + CO2 + Na2SO4 + H20 half-reaction for the oxidation: 204 + H2S04 Mnso. Consider the fol experiment: (PI KMNO4 + Na2C2a a. Write a balanced. ced, unmultiplied lonic half-react ed, unmultiplied ionic half-reactione alf-reaction for the reduction: b. Write a balanced, unm e balanced net-ionic reaction: C. Write the balanced the balanced molecular reaction: d. Write the balanced 2. How many mL of a 1.356 NK solution are needed to completely react with in acid solution. Show all work and units. titrate) 2.500 grams of Na2C204 in acid solutio

Answer 1

1a. C₂O₄^2- is oxidized to CO₂ (ignore counter ions since they do no participate in the reaction).

The reaction takes place in acid medium; hence, H⁺ and H₂O are present in the system.

C₂O₄^2- (aq) ----------> 2 CO₂ (g) + 2 e^-

1b. MnO₄^- is reduced to Mn²⁺ in acidic medium.

MnO₄^- (aq) + 8 H⁺ (aq) + 5 e^- ---------> Mn²⁺ (aq) + 4 H₂O (l)

1c. In order to balance the ionic equations, multiply the oxidation half by 5 and the reduction half by 2 and add.

5 C₂O₄^2- (aq) + 2 MnO₄^- (aq) + 16 H⁺ (aq) ----------> 10 CO₂ (g) + 2 Mn²⁺ (aq) + 8 H₂O (l)

1d. The molecular equation can be obtained by adding the counter ions and noting that 16 H⁺ must come from 8 H₂SO₄.

5 Na₂C₂O₄ (aq) + 2 KMnO₄ (aq) + 8 H₂SO₄ (aq) --------> 10 CO₂ (g) + 2 MnSO₄ (aq) + K₂SO₄ (aq) + 5 Na₂SO₄ (aq) + 8 H₂O (l)

2. Normality of KMnO₄ = 1.356 N

The normality and molarity of KMnO₄ are related as

Molarity = 5*(normality)

=====> Molarity = 5*(1.356 N) = 6.78 M

The concentration of KMnO₄ = 6.78 M.

The atomic masses are

Na: 22.989 u

C: 12.011 u

O: 15.999 u

Gram molar mass of Na₂C₂O₄ = (2*22.989 + 2*12.011 + 4*15.999) g/mol

= 133.996 g/mol.

Mole(s) of Na₂C₂O₄ corresponding to 2.500 g = (2.500 g)/(133.996 g/mol)

= 0.01866 mole.

As per the stoichiometric equation in 1d above,

5 moles Na₂C₂O₄ = 2 moles KMnO₄.

Therefore,

0.01866 mole Na₂C₂O₄ = (0.01866 mole Na₂C₂O₄)*(2 moles KMnO₄)/(5 moles Na₂C₂O₄)

= 0.007464 mole KMnO₄.

Volume of 6.78 M KMnO₄ required = (0.007464 mole)/(6.78 M)

= (0.007464 mole)/(6.78 mol/L)

= 0.001101 L

= (0.001101 L)*(1000 mL)/(1 L)

= 1.101 mL

≈ 1.1 mL (ans).