In a volumetric analysis experiment,an acidic aqueous solution of methanol (CH₃OH) is titrated with a solution of potassium dichromate (K₂Cr₂O₇) according to the following balanced chemical equation:
2K₂Cr₂O₇(aq) + 8H₂SO₄(aq) + 3CH₃OH(aq) → 2Cr₂(SO₄) ₃(aq) + 11H₂O(l) + 3HCOOH(aq) + 2K₂SO₄(aq)
It required 49.73 mL of 0.0153 M K₂Cr₂O₇ to reach the endpoint.What mass of CH₃OH was present initially?

Question

Quizplus · Accepted Answer

First, calculate the moles of $K_2Cr_2O_7$ used: $49.73 \, \text{mL} \times 0.0153 \, \text{M} = 0.000761 \, \text{mol}$. From the balanced equation, 2 moles of $K_2Cr_2O_7$ react with 3 moles of $CH_3OH$, so the moles of $CH_3OH$ are $0.000761 \, \text{mol} \times \frac{3}{2} = 0.0011415 \, \text{mol}$. The molar mass of $CH_3OH$ is approximately 32 g/mol, so the mass of $CH_3OH$ is $0.0011415 \, \text{mol} \times 32 \, \text{g/mol} = 0.0365 \, \text{g}$.

In a Volumetric Analysis Experiment,an Acidic Aqueous Solution of Methanol