Step 1: Determine the limiting reagent.
I'm guessing water is in excess since a mass is not given, so #C_12H_22O_11# is our limiting reagent.
Step 2: Convert grams into moles.
#"68 grams" C_12H_22O_11*(("1 mole" C_12H_22O_11)/("342.297 grams"))="0.20 moles" C_12H_22O_11#
Step 3: Use the mole ratio between #C_12H_22O_11# and #C_2H_5OH# to find how many moles of product will be produced.
The mole ratio between the two is 1 mole of #C_12H_22O_11# for every 4 moles of #C_2H_5OH#, therefore for every 1 mole of #C_12H_22O_11# we react, we will get 4 moles of #C_2H_5OH#.
To do this for 0.20 moles, we do the stoicheometry:
#"0.20 moles" C_12H_22O_11*(("4 moles" C_2H_5OH)/("1 mole" C_12H_22O_11))="0.80 moles" C_2H_5OH#
This number is the moles of #C_2H_5OH# produced per 0.20 moles of #C_12H_22O_11#.
Step 4: Calculate the mass of the #C_2H_5OH#
#"0.80 moles" C_2H_5OH*(("46.069 grams")/("1 mole" C_2H_5OH))="37 grams" C_2H_5OH#
This number is our theoretical yield of #C_2H_5OH# in grams for every 0.20 moles of #C_12H_22O_11#
Step 5: Calculate percent yield.
The equation for percent yield is:
#"% yield"=("actual yield"/"theoretical yield")*100#
So we can plug in the numbers:
#"% yield"=(31.2/37)*100=84%#
I hope this helps!
