Chapter 23, Problem 100A

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the synthesis of ethanol from ethene and water is to be written.

Concept introduction:

Balanced chemical equation is the chemical equation in which the number of atoms of every element in both sides of chemical reaction, that is, product side and reactant side.

Answer to Problem 100A

The balanced equation for the synthesis of ethanol from ethene and water is,

${\text{C}}_{\text{2}}{\text{H}}_4(g)+{\text{H}}_{\text{2}}\text{O}\text{(}l\text{)}\to {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$

Explanation of Solution

The reaction of ethene and water produces ethanol.

${\text{C}}_{\text{2}}{\text{H}}_4(g)+{\text{H}}_{\text{2}}\text{O}\text{(}l\text{)}\to {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$

The above equation is balanced chemical equation as the number of atoms of carbon, hydrogen and oxygen is equal in both sides of the reaction.

(b)

Interpretation Introduction

Interpretation:

The amount of ethanol in grams is produced on reaction of 448 L of ethene gas with excess water at STP needs to be calculated.

Concept introduction:

The Standard temperature and pressure, STP is when standard pressure is 1 atm and standard temperature is 273 K $\left(0°\text{C}\right)$. 1 mole of any gas at STP condition has a volume volume equal to 22.4 Litre.

Answer to Problem 100A

The amount of ethanol is produced on reaction of 448 L of ethene gas with excess water at STP is 921 g.

Explanation of Solution

The amount of ethanol (in grams) produced can be calculated by dividing the volume of ethene by standard volume for one mole of ethene, that is, 22.4 Litres. Then, multiply the product by the mole ratio between ethanol and ethene, which is 1:1, respectively. At last, multiply the solution by the molar mass of ethanol, which is 46.07 g/mol.

$\begin{array}{c}\text{Mass}\text{of}\text{ethanol}\text{=448}\text{L}\text{ethene}\times \frac{\text{1}\text{mol}\text{ethene}}{22.4\text{L}\text{ethene}}\times \frac{1\text{mole}\text{ethanol}}{1\text{mole}\text{ethene}}\times \frac{46.07\text{g}\text{ethanol}}{1\text{mole}\text{ethanol}}\\ =921\text{g}\end{array}$