Question

The combustion of liquid ethanol (C2H5OH) produces carbon dioxide and water. After 4.61 mL of ethanol (density=0.789g/ml) was allowed to burn in the presence of 15.70 g of oxygen gas, 3.71 mL of water (density=1.00g/ml) was collected. Determine the theoretical yield of H2O for the reaction. Determine the percent yield of H2O for the reaction.

Answer 1

Solution: C2H5OH(l) + 3O2(g) ----> 2CO2(g) + 3H2O(l)

Density of ethanol=0.789 g

mass of ethanol = 0.789*4.61=3.64 grams

mass of O2=15.70 grams

calculating the moles of ethanol & moles of O2

moles of ethanol=3.64/46.1=0.0789

moles of O2= 15.70/32=0.49

determining the limiting reagent.

from the balanced equation

1 mole of ethanol reacts with 3 moles of O2

0.0789 moles ethanol produces ? moles of O2

=> moles of O2 =(0.0789*3)/1=>0.2367 moles of O2

since given moles of O2(i.e 0.49 moles ) is greater then required moles of O2(i.e 0.2367 moles)O2 is not the limiting reagent.

hence ethanol is the limiting reagent

from the balanced equation:

1 mole of ethanol produces 3 moles of H2O

0.0789moles of ethanol produces? moles of H2O

> (0.0789 *3)/1=0.2367 moles of H2O

mass of water = 0.2367*18= 4.3 grams of water

volume of water= 4.3ml

theoritical yield of water= 4.3 ml

percent yield= (3.71/4.3)*100= 86.3 %

percent yield= 86.3 %