Question #daeef

1 Answer
Dec 4, 2017

See below.

Explanation:

Step 1: Determine the limiting reagent.

I'm guessing water is in excess since a mass is not given, so C12H22O11 is our limiting reagent.

Step 2: Convert grams into moles.

68 gramsC12H22O11(1 moleC12H22O11342.297 grams)=0.20 molesC12H22O11

Step 3: Use the mole ratio between C12H22O11 and C2H5OH to find how many moles of product will be produced.

The mole ratio between the two is 1 mole of C12H22O11 for every 4 moles of C2H5OH, therefore for every 1 mole of C12H22O11 we react, we will get 4 moles of C2H5OH.

To do this for 0.20 moles, we do the stoicheometry:

0.20 molesC12H22O11(4 molesC2H5OH1 moleC12H22O11)=0.80 molesC2H5OH

This number is the moles of C2H5OH produced per 0.20 moles of C12H22O11.

Step 4: Calculate the mass of the C2H5OH

0.80 molesC2H5OH(46.069 grams1 moleC2H5OH)=37 gramsC2H5OH

This number is our theoretical yield of C2H5OH in grams for every 0.20 moles of C12H22O11

Step 5: Calculate percent yield.

The equation for percent yield is:

% yield=(actual yieldtheoretical yield)100

So we can plug in the numbers:

% yield=(31.237)100=84%

I hope this helps!