Molar mass?

1.

(i)

Percentage of C in the molecule = (171.36/200) × 100% = 85.68%

Percentage of N in the molecule = (18.18/200) × 100% = 9.09%

Percentage of H in the molecule = [(100 - 85.68 - 9.09)% = 5.23%

(ii)

In each mole of the acid:

No. of moles of C = (616.73 × 85.68%) / 12.0 = 44

No. of moles of H = (616.73 × 5.23%) / 1.0 = 32

No. of moles of N = (616.73 × 9.09%) / 14.0 = 4

Molecular formula for the acid = C₄₄H₃₂N₄

(ii)

(200 g C₄₄H₃₂N₄) × (1 mol C₄₄H₃₂N₄ / 616.73 g) × (6.022 × 10²³ molecules / 1 mol) × (4 N atoms / 1 molecule)

= 7.81 × 10²³ N atoms

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2.

(i)

NaCrO₄ does not exist. This is because the oxidation number of Cr in NaCrO₄ is +7, but the maximum oxidation number of Cr is +6.

Suppose it is "Na₂CrO₄" instead. The equation would be:

2CrCl₃ + 3Na₂O₂ + 4NaOH → 2Na₂CrO₄ + 6NaCl + 2H₂O

(ii)

Oxidation half equation: Na₂O₂(s) + 2H₂O(l) + 2e⁻ → 2Na⁺(aq) + 4OH⁻(aq)

Reduction half equation: Cr₃⁺(aq) + 8OH⁻(aq) → CrO₄²⁻(aq) + 4H₂O(l) + 3e⁻

(iii)

Na₂O₂ is reduced, and CrCl₃ (or Cr³⁺) is oxidized.

(iv)

The greatest common factor of 2 and 3 is 6. Hence, 6 electrons participate in the redox process.

Large molecular weight acid....

MESO-TETRAPHENYLCHLORIN

https://www.tcichemicals.com/image/commodity/T1358...

.... among other possible compounds....

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Note: there is in an error in the question. It should be Na2CrO4 on the product side.

Redox reaction....

2(Cr3+ + 8OH- --> CrO4^2- + 4H2O + 3e-) ........... Oxidation

3(O2^2- + 2H2O + 2e- --> 4OH-) ................ .......... Reduction

------------------ -------------------- ------------------- ---------------------

2Cr3+ + 16OH- + 3O2^2- + 6H2O --> 2CrO4^2- + 8H2O + 12OH-

simplify

2Cr3+ + 4OH- + 3O2^2- --> 2CrO4^2- + 2H2O .... net ionic equation

Chromium (as Cr3+) is oxidized, oxygen (as peroxide) is reduced.

Six electrons are "transferred."

Wouldn't it be easier to read if the photo was rotated properly?

.

%C = 171.36g / 200g =85.68%

C per mole =528 g or 44 moles of C

%N = 18.18g / 200g = 9.09%

N per mole = 56.0 g = 4 mole of N

%H= 5.23% = 32 g of H and 32 moles of H

molecular formula is C44H32N4

in 200 g of this compound there are 56 . 0 g or 4 moles of N or 4 * 6,02 *10^23 atoms of N

it cant be an acid as there is no Oxygen figure quoted