For the oxidation of acetaldehyde (a) moles acetaldehyde = 0.3250 g/44.054 g/mol

Question

For the oxidation of acetaldehyde

(a) moles acetaldehyde = 0.3250 g/44.054 g/mol = 0.0074 mol

moles HCl = 1.250 M x 0.045 L = 0.05625 mol

moles Na2Cr2O7 = 0.7776 g/262 g/mol = 0.003 mol

For complete consumption of acetaldehyde we need = 0.0074/3 = 0.0025 mol Na2Cr2O7

For complete consumption of acetaldehyde we need = 0.0074 x 8/3 = 0.02 mol HCl

Since moles of Na2Cr2O7 and HCl are much higher then needed, acetaldehyde is the limiting reactant

(b) In the final solution,

we would have, Na+, Cr2O7^2-, aceataldehyde, H3O+, Cl- will be found in the final reaction solution

(c) In the final solution (after the reaction)

[Na+] = 0.003 mol x 2/0.045 L = 0.1333 M

[Cr2O7^2-] = 0.0005 mol/0.045 L = 0.0111 M

[Cl-] = 0.03625 mol/0.045 L = 0.8055 M

[H3O+] = 0.03625 mol/0.045 L = 0.8055 M

Explanation / Answer

3. Acetaldehyde, CoH,O, is an organic compound that can be oxidized by acidic dichromate solutions to give acetic acid. HC2H,O2. The net ionic equation describing this reaction is: Consider adding 0.3250 grams of acetaldehyde to 45.0 mL of an aqueous acidic solution which is 1.250M in HCI and contains 0.7776 grams of sodium dichromate. Assume the above reaction takes place. (a) What is the limiting reagent? Be sure you use the mol-to-mol ratio in finding both the limiting reagent and the amounts of products and amounts of unused reactants. (b) What reagents (both neutral species and ions) will be found in the final solution? (c) What ions will be found in the final solution and calculate the molarities of each of these ions. The following formula weights may be used if needed: HCI (86.53 amul; CH40 (44.054 amu), and Na Cr20, (262 amu); HC,H,02 (60.05 amu).

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