Calculate the volumes of benzaldehyde and acetone needed for this reaction. You

Question

Calculate the volumes of benzaldehyde and acetone needed for this reaction. You mustdetermine the mass of each reactant needed using the stoichiometry of the reaction and then convert the masses to volumes using the density of each compound.

The procedure starts off with "Mix 0.05 mole of benzaldehyde with the stoichiometric quantity of acetone in a small beaker"

I calculated the benzaldehyde but I'm not sure if I got it right.

0.05 mol x 106.13 g/mol = 5.3065 g / 1.04 g/mL = 5.10 mL

For acetone:
0.05 mol x 58.08 g/mol = 2.904 g / 0.790 g/ mL = 3.68 mL

However, I'm a little confused. One of my classmates got 1.15 mL of acetone for the answer. If that's not right, how would you calculate it?

Explanation / Answer

from the given equation,

2 mol benzaldehyde = 1 mol acetone

so that,

0.05 mol benzaldehyde = 0.025 mol acetone

mass of benzaldehyde = 0.05*106.13 = 5.3065 grams

volume of benzaldehyde = 5.3065/1.04 = 5.1 ml

similarly for acetone.

mass of acetone = 0.025*58.08 = 1.452 grams

volume of acetone = 1.452/0.79 = 1.838 ml

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