[imari]

I have two questions that I am struggling with.

1. When 50.0 mL of a certain Na₂CO₃ solution was combined with 0.102 M HCl, 56.3 mL of HCl were required for complete neutralization, according to the equation:

                             CO₃^2- + 2H⁺ --> CO₂ +H₂O

Calculate the number of grams of CaCO₃ that would be precipitated if an excess of CaCl₂ were added to a separate 50.0 mL portion of the Na₂CO₃ solution.


I did work for this and got an answer but I am not sure if it is correct:

0.102 mol/L * 0.0563 L = 0.00574 mol HCL

0.00574 mol H * 1/2 = 0.00287 mol CO₃

Since Ca²⁺ + CO₃^2- --> CaCO₃

0.00287 mol CaCO₃ * 100 g/mol = 0.287g CaCO₃



2. Balance the following oxidation-reduction equations:

a) Zn + NaNO₃ + NaOH --> Na₂ZnO₂ + NH₃ + H₂O

First I split them up into two half reactions and then balanced them but the electrons and charges confused me so i am not sure if i did it right.

oxidation: Zn + 2NaOH --> Na₂ZnO₂ + H⁺ + 2e^-
reduction: 2e^- + 9H⁺ + NaNO₃ --> NH₃ + 3H₂O + Na₊

Then I added both the right sides and then both the left sides, without writing the two electrons on both sides. Are the half reactions correct?

b) MnSO₄ + (NH₄)₂S₂O₈ + H₂O --> MnO₂ + H₂SO₄ + (NH₄)₂SO₄

It looks to me that the Mn is oxidized and the S is reduced but when I try to write the half reactions, I am having trouble balancing them. How do I write the half reactions?

[Borek]

0.00287 mol CaCO₃ * 100 g/mol = 0.287g CaCO₃

Correct

2. Balance the following oxidation-reduction equations:

First of all - write them in net ionic form, this way it is easier to concentrate on what is going on.

[AWK]

1. OK
2

oxidation: Zn + 2NaOH --> Na2ZnO2 + H+ + 2e-
reduction: 2e- + 9H+ + NaNO3 --> NH3 + 3H2O + Na+

Be more careful when balancing half reaction - atoms and cherges should be balanced
oxidation: Zn + 2NaOH --> Na₂ZnO₂ + 2H⁺ + 2e^-
reduction: 8e^- + 9H⁺ + NaNO₃ --> NH₃ + 3H₂O + Na⁺
Show your attempt for 3