Q. 7.111 - An Introduction to General, Organic, and Biological Chemistry

Question

Consider the following unbalanced equation: ($7.4, 7.5, 7.7, 7.8$)

$\mathrm{Al}\left(\mathrm{s}\right) + {\mathrm{O}}_2\left(\mathrm{g}\right) \to {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$

a. Write the balanced chemical equation. 

b. Identify the type of reaction.

c. How many moles of $O_2$ are needed to react with $4.50$ mole of $\mathrm{Al}$?

d. How many grams of ${\mathrm{Al}}_2{\mathrm{O}}_3$ are produced when $50.2$ g of $\mathrm{Al}$ reacts?

e. When $\mathrm{Al}$ is reacted with $8.00$ g  of ${\mathrm{O}}_2$, how many grams of $A{l}_2{O}_3$ can form?

Step-by-Step Solution

Answer

(a) The required balanced equation is $4\mathrm{Al}+3{\mathrm{O}}_2\to 2{\mathrm{Al}}_2{\mathrm{O}}_3$

(b) The given reaction is a combination reaction

(c) The $3.38$ moles of oxygen are needed to react with $4.50$ moles of aluminum

(d)The $94.9$ g of ${\mathrm{Al}}_2{\mathrm{O}}_3$ are produced when $50.2$ g of aluminum reacts.

(e) The $17$ g of ${\mathrm{Al}}_2{\mathrm{O}}_3$ forms when $8$ g of oxygen reacts with aluminium

Part (a) Step 1: Given information

We need to find the balance the given equation

Part (a) Step 2: Explanation

We know that

A balanced equation is that in which total energy of reactants is equal to total energy of products.

Therefore, the required balanced equation is, 

$4\mathrm{Al}+3{\mathrm{O}}_2\to 2{\mathrm{Al}}_2{\mathrm{O}}_3$

Part (b) Step 1: Given information

We need to find the type of the given equation

Part (b) Step 2: Explanation

As in the given equation, aluminum and oxygen react with each other to form a single compound which is characteristic of a combination reaction

Therefore, The given reaction is a combination reaction 

Part (c) Step 1: Given information

We need to find the no. of moles of oxygen that are needed to react with $4.50$ mole of aluminum 

Part (c) Step 2: Explanation

From part (a)

We know that

The $4$ mole of aluminum reacts with $3$ mole of oxygen

Therefore, no. of moles of oxygen reacting with $4.5$ mole of aluminum are, $\frac{3 \cancel{\mathrm{mole}}}{4 \cancel{\mathrm{mole}}}\times 4.5 \mathrm{mole}=3.38 \mathrm{mole}$

Part (d) Step 1: Given information

We need to find amount of $A{l}_2{O}_3$ produced from $50.2$ g of aluminum

Part (d) Step 2: Explanation

From part (a) 

We know that

The $108$ g of aluminum produces $204$ g of ${\mathrm{Al}}_2{\mathrm{O}}_3$

Therefore, $50.2$ g of aluminum produces $\frac{204 \cancel{\mathrm{g}}}{108 \cancel{\mathrm{g}}}\times 50.2 \mathrm{g}=94.9 \mathrm{g}$ of ${\mathrm{Al}}_2{\mathrm{O}}_3$

Part (e) Step 1: Given information

We need to find the amount of ${\mathrm{Al}}_2{\mathrm{O}}_3$ formed when $8$ of oxygen reacts with aluminum

Part (e) Step 2: Explanation

From part (a) 

We know that

The $96$ g of oxygen forms $204$ g of ${\mathrm{Al}}_2{\mathrm{O}}_3$

Therefore, $8$ g of oxygen forms $\frac{204 \cancel{\mathrm{g}}}{96 \cancel{\mathrm{g}}}\times 8 \mathrm{g}=17 \mathrm{g}$ of ${\mathrm{Al}}_2{\mathrm{O}}_3$