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 localid="1653472797651" $72$gof water?

c. How many grams of ${\mathrm{CO}}_2$are produced from localid="1653472811626" $32.0$gof ${\mathrm{O}}_2$?

Short 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 localid="1653472833201" $48$gof pentane gas is needed to produce localid="1653472847302" $72$gof water

(c) The localid="1653472861523" $27.5$gof ${\mathrm{CO}}_2$are produced from localid="1653472874083" $32$gof ${\mathrm{O}}_2$

Step-by-step solution

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 $72$gof water

Part (b) Step 2: Explanation

From part (a)

We know that

The $108$gof water is produced by $72$gof pentane gas

Therefore,$72$gof 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 $32$gof $O_2$

Part (c) Step 2: Explanation  

From part (a)

We know that

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

Therefore, localid="1653473147345" $32$gof${\mathrm{O}}_2$produceslocalid="1653485137564" $\frac{220\cancel{\mathrm{g}}}{256\cancel{\mathrm{g}}}\times 32\mathrm{g}=27.5\mathrm{g}$of${\mathrm{CO}}_2$