Question

Ethanol reacts with sodium dichromate in an aqueous hydrochloric acid solution according to the balanced net ionic equation shown below.

3 C₂H₅OH (aq) + 2 Cr₂O₇^2- (aq) + 16 H⁺ (aq) → 3 HC₂H₃O₂ (aq) + 4 Cr³⁺ (aq) + 11 H₂O (l)

This reaction is carried out with 1.00 mL of C₂H₅OH (density 0.816 g/mL, molar mass 46.07 g/mole) reacting with 32.76 mL of a solution that is 0.5775 M in Na₂Cr₂O₇ and 6.000 M in HCl.

What is the limiting reagent?

The mass of CrCl₃ (molar mass 158.36 g/mole) formed? (in grams)

Answer 1

Mass of ethanol in 1.00 mL solution = density x volume = 0.816 g/mL x 1.00 mL = 0.816 g

Molar mass of ethanol = 46.07 g/mol

Moles of ethanol (C₂H₅OH) = mass of ethanol/molar mass of ethanol

                                             = 0.816 g /46.07 g/mol

                                             = 0.0177 mol

Volume of Na₂Cr₂O₇ and HCl solution = 32.76 mL = 0.03276 L    ( 1 mL = 0.001 L)

Moles of Na₂Cr₂O₇ = 0.03276 L x 0.5775 M = 0.01892 mol

Moles of HCl = 0.03276 L x 0.6000 M = 0.01966 mol

Mole ratio of C₂H₅OH : Na₂Cr₂O₇ : HCl = 0.0177 : 0.01892 : 0.01966

                                                                = 1.0 : 1.1 : 1.1        (dividing all by 0.0177)

                                                                = 3.0 : 3.3 : 3.3        (multiplying all by 3)

From the balanced equation given in the problem, it is clear that 16 mol of H⁺ or HCl is required completely with 3 mol of ethanol. But here, only 3.3 mol of HCl is available when 3 mol of ethanol is used.

Therefore, HCl is the limiting reagent for this reaction.

Now, theoretically, 16 mol of HCl gives 4 mol of Cr³⁺ or CrCl₃

Therefore, 0.01966 mol of HCl gives = 4 mol x 0.01966 mol/16 mol = 0.004915 mol of CrCl₃

Now, molar mass of CrCl₃ = 158.36 g/mol

Hence, the mass of CrCl₃ formed = 0.004915 mol x 158.36 g/mol

                                                      = 0.7783 g