Q. 7.105 - An Introduction to General, Organic, and Biological Chemistry

Question

Pentane gas, ${\mathrm{C}}_5{\mathrm{H}}_{12}$, undergoes combustion with oxygen gas to produce carbon dioxide and water gases. ($7.4, 7.8$)

a. Write the balanced chemical equation. 

b. How many grams of ${\mathrm{C}}_5{\mathrm{H}}_{12}$ are needed to produce $72$ g of water?

c. How many grams of ${\mathrm{CO}}_2$ are produced from $32.0$ g of ${\mathrm{O}}_2$?

Step-by-Step Solution

Answer

(a) The required balanced equation is ${\mathrm{C}}_5{\mathrm{H}}_{12}+8{\mathrm{O}}_2\stackrel{∆}{\to }5{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}+\mathrm{energy}$

(b) The $48$ g of pentane gas is needed to produce $72$ g of water

(c) The $27.5$ g of ${\mathrm{CO}}_2$ are produced from $32$ g of ${\mathrm{O}}_2$

Part (a) Step 1: Given information

We need to find the balanced equation

Part (a) Step 2: Explanation

We know that

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

Therefore, The balanced equation is, 

${\mathrm{C}}_5{\mathrm{H}}_{12}+8{\mathrm{O}}_2\stackrel{∆}{\to } 5{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}+\mathrm{energy}$

Part (b) Step 1: Given information

We need to find the mass of pentane gas needed to produce $72$ g of water 

Part (b) Step 2: Explanation

From part (a)

We know that

The $108$ g of  water is produced by $72$ g of pentane gas

Therefore, $72$ g of water is produced by $\frac{72 \cancel{\mathrm{g}}}{108 \cancel{\mathrm{g}}}\times 72 \mathrm{g}=48 \mathrm{g}$ of pentane gas 

Part (c) Step 1: Given information

We need to find mass of ${\mathrm{CO}}_2$ produced from $32$ g of $O_2$

Part (c) Step 2: Explanation

From part (a)

We know that

The $256$ g of ${\mathrm{O}}_2$ produces $220$ of ${\mathrm{CO}}_2$

Therefore, $32$ g of ${\mathrm{O}}_2$ produces $\frac{220 \cancel{\mathrm{g}}}{256 \cancel{\mathrm{g}}}\times 32 \mathrm{g}=27.5 \mathrm{g}$ of ${\mathrm{CO}}_2$