Q. 7.106 - An Introduction to General, Organic, and Biological Chemistry

Question

Propane gas, ${\mathrm{C}}_3{\mathrm{H}}_8$, undergoes combustion with oxygen gas to produce carbon dioxide and water gases. Propane has a density of $2.02 g/L$ at room temperature. ($7.4, 7.8$)

a. Write the balanced chemical equation.

b. How many grams of ${\mathrm{H}}_2\mathrm{O}$ form when $5.00 L$ of ${\mathrm{C}}_3{\mathrm{H}}_8$ reacts?

c. How many grams of ${\mathrm{H}}_2\mathrm{O}$ can be produced from the reaction of $100 g$ of ${\mathrm{C}}_3{\mathrm{H}}_8$?

Step-by-Step Solution

Answer

(a) The required balanced equation is ${\mathrm{C}}_3{\mathrm{H}}_8+5{\mathrm{O}}_2\to 3{\mathrm{CO}}_2+4{\mathrm{H}}_2\mathrm{O}$

(b) The $16.5 g$ of ${\mathrm{H}}_2\mathrm{O}$ form when $5.00 L$ of ${\mathrm{C}}_3{\mathrm{H}}_8$ reacts

(c) The $163.6 g$ of ${\mathrm{H}}_2\mathrm{O}$ can be produced from the reaction of $100 g$ of ${\mathrm{C}}_3{\mathrm{H}}_8$

Part (a) Step 1: Given information

We need to find the balanced chemical equation

Part (a) Step 2: Explanation

We know that

A balanced equation is that in which total mass of the reactants is equal to the total mass of products

Therefore, the required balanced equation is,

${\mathrm{C}}_3{\mathrm{H}}_8+5{\mathrm{O}}_2\to 3{\mathrm{CO}}_2+4{\mathrm{H}}_2\mathrm{O}$

Part (b) Step 1: Given information

We need to find the amount of ${\mathrm{H}}_2\mathrm{O}$ form when $5.00 L$ of data-custom-editor="chemistry" ${\mathrm{C}}_3{\mathrm{H}}_8$ reacts

Part (b) Step 2: Explanation

We know that

The mass of ${\mathrm{C}}_3{\mathrm{H}}_8$ in $5.00 L$ of propane is equal to the multiplication  of density and volume

Hence, the mass of propane is, $2.02\times 5=10.1 g$

And from part (a) 

We know that  $44 g$ of propane gas forms  $72 g$ of water

Therefore, $10.1 g$ of propane gas forms $\frac{72}{44}\times 10.1=16.5 g$ of water

Part (c) Step 1: Given information

We need to find the amount of ${\mathrm{H}}_2\mathrm{O}$ can be produced from the reaction of $100 g$ of ${\mathrm{C}}_3{\mathrm{H}}_8$

Part (c) Step 2: Explanation

From part (a)

We know that

The $44 g$ of propane forms $72 g$ of water

Therefore, $100 g$ of propane forms $\frac{72}{44}\times 100=163.6 g$ of water