Chapter 11, Problem 90A

Interpretation Introduction

Interpretation : The theoretical and percent yield of fermentation of sugar needs to be determined.

Concept introduction : On fermentation of sugar in presence of enzyme ethanol is produced. It is the main source for production of ethanol.

Answer to Problem 90A

Theoretical Yield = 368 g

Percent Yield = 94.83%

Explanation of Solution

Given amount of sugar = 684 g

Ethanol produced from fermentation (Actual yield) = 349 g

To determine theoretical yield:

Step 1: write balance chemical equation

$\begin{array}{l}\text{C12H22O11(aq) + H2O(l)}\to \text{ 4C2H5OH(l) + 4CO2(g)}\\ \text{ 648g 349g}\end{array}$

Step 2: Convert to mole

Molar mass of sugar = 342 g / mol

$\begin{array}{c}\text{Moles = }\frac{Mass}{Molarmass}\\ =\frac{684}{342}=2\end{array}$

Step 3: use mole ratio to convert moles of reactant to product.

From reaction stoichiometry

1 mole of sugar produce = 4 mole of ethanol

2 mole of sugar produce = 4 X 2 moles of ethanol = 8 moles of ethanol

Step 4: Convert moles of product to mass (theoretical yield)

Molar mass of ethanol = 46 g/mol

$\begin{array}{c}Mass=Molarmass\times Moles\\ =46\times 8\\ =368g\end{array}$

Percent Yield is to be calculated using relation given below:

$\begin{array}{c}PercentYield=\frac{Actualyield}{Theoreticalyield}\times 100\\ =\frac{349}{368}\times 100=94.83\%\end{array}$