Question

The balanced the equation

1 CH3C(O)CH3 + 2 CH3CH2OH +sulfuric acid (catalyst)->1 (CH3)2 C(OCH2CH3)2 + H2O

Assume 2.00g acetone (CH3C(O)CH3) is reacted with 3.60g ethanol in the presence of 0.400g sulfuric acid catalyst to produce 1.00g of the product acetal. What is the limiting reagent?

Answer 1

We need to compare moles of species

assume reaciton is balanced

mol of Acetone = mass of Acetone / MW of Acetone = 2/58.08 = 0.0344 mol of acetone

mol of ethanol = mass/MW = 3.6 / 46.06844 = 0.0781 mol of ethanol

mol of acid = mass/MW = 0.4/98 = 0.004

What is the limiting reagent?

0.0344 mol of acetone requires:

2x0.0344 = 0.0688 mol of ethanol

we have

0.0781 mol of ethanol; therefore, ethanol is in excess = 0.0781 -0.0688 = 0.0093

therefore

acetone is limiting reaction