1C6H12O6(aq)---->2C2H5OH(l)+2CO2(g)

Question

If  350.0 g of glucose yields 103.0 mL of ethanol, what is the percent yield for the reaction?

Answer 1

C6H12O6(aq) ----> 2C2H5OH(l) + 2CO2(g)

glucose .................ethanol............carbon dioxide

180.16g/mol.........46.07g/mol....................

From the balanced equation above, we see that ONE mole glucose produces 2 moles of ethanol

molar mass glucose = 180.16 g/mol

moles glucose used = 350.0 g x 1 mol / 180.16 g = 1.9427 moles

Theoretical yield of ethanol = 1.9427 moles glucose x 2 mol ethanol / mole glucose = 3.8854 mols ethanol

Theoretical yield of ethanol in grams = 3.885 mols x 46.07 g/mol = 178.98 g

Actual yield of ethanol = 103.0 mls.

The density of ethanol = 0.789 g/ml @20ºC (since no other information is given in the problem, we'll use this)

mass of ethanol produced = 103.0 ml x 0.789 g/ml = 81.27 g

%yield of ethanol = actual/theoretical (x100%)

%yield of ethanol = 81.27g / 178.98 g (x100%) = 45.41% yield