Q. 7.110 - An Introduction to General, Organic, and Biological Chemistry

Question

Gasohol is a fuel containing liquid ethanol (${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$) that burns in oxygen gas to give carbon dioxide and water gases. ($7.4, 7.7, 7.8$)

a. Write the balanced chemical equation.

b. How many moles of ${\mathrm{O}}_2$ are needed to completely react with style="max-width: none; vertical-align: -4px;" $4.0$ mole of ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$?

c. If a car produces style="max-width: none; vertical-align: -4px;" $88$ g of ${\mathrm{CO}}_2$, how many grams of ${\mathrm{O}}_2$ are used up in the reaction?

d. If you burn $125$ g of ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$, how many grams of ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$ can be produced?

Step-by-Step Solution

Answer

(a) The balanced chemical equation is ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}+3{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}$

(b) The style="max-width: none; vertical-align: -4px;" $12$ mole of ${\mathrm{O}}_2$ needed to completely react with style="max-width: none; vertical-align: -4px;" $4.0$ mole of ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ 

(c) The $96$ g of ${\mathrm{O}}_2$ are used up in the reaction

(d) The $239$ g of ${\mathrm{CO}}_2$ and $146.7$ g of ${\mathrm{H}}_2\mathrm{O}$ can be produced

Part (a) Step 1: Given information

We need to find the balanced chemical equation

Part (a) Step 2: Explanation

We know that

The balanced equation is that in which the total mass of reactant is equal to the total mass of products

Therefore, The required balanced equation is,

${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}+3{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}$

Part (b) Step 1: Given information

We need to find the no. of moles of ${\mathrm{O}}_2$ are needed to completely react with $4.0$ mole of ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ 

Part (b) Step 2: Explanation

From part (a) 

We know that

A mole of ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ reacts with $3$ mole of ${\mathrm{O}}_2$

Therefore, $4$ mole of ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ reacts with $\frac{3 \mathrm{mole}\times 4 \cancel{\mathrm{mole}}}{\cancel{\mathrm{mole}}}=12 \mathrm{mole}$ of ${\mathrm{O}}_2$

Part (c) Step 1: Given information

We need to find the mass of ${\mathrm{O}}_2$ is used up in the reaction

Part (c) Step 2: Explanation

From part (a) 

We know that 

The $88$ g of ${\mathrm{CO}}_2$ is produced from $96$ g of ${\mathrm{O}}_2$

Therefore, The $96$ g of ${\mathrm{O}}_2$ are used up in the reaction

Part (d) Step 1: Given information

We need to find the amount of ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$ produced 

Part (d) Step 2: Explanation

From part (a) 

We know that 

The $46$ g of ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ produces $88$ g of ${\mathrm{CO}}_2$ and $54$ g of ${\mathrm{H}}_2\mathrm{O}$

Therefore, $125$ g of ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ produces $\frac{88 \cancel{\mathrm{g}}}{46 \cancel{\mathrm{g}}}\times 125 \mathrm{g}=239 \mathrm{g}$ of ${\mathrm{CO}}_2$ and $\frac{54 \cancel{\mathrm{g}}}{46 \cancel{\mathrm{g}}}\times 125 \mathrm{g}=146.7 \mathrm{g}$ of ${\mathrm{H}}_2\mathrm{O}$