Q. 11.72 - An Introduction to General, Organic, and Biological Chemistry

Question

Consider the compound ethylcyclopentane. $(7.4,7.7,8.6,11.3,11.4)$

a. Draw the line-angle formula for ethylcyclopentane.

b. Write the balanced chemical equation for the complete combustion of ethylcyclopentane.

c. Calculate the grams of $O_2$ required for the combustion of$25.0g$ of ethylcyclopentane.

d. How many liters of $C{O}_2$ would be produced at STP from the reaction in part c?

Step-by-Step Solution

Answer

(a) The line-angle formula for ethylcyclopentane is

(b) The balanced chemical equation for the complete combustion of ethylcyclopentane is

$2C_7{H}_{14}+21O_2\to 14C{O}_2+14H_{2}O$.

(c) $85.71g$of $O_2$required for the combustion of $25.0g$of ethylcyclopentane.

(d) $40L$of $C{O}_2$ would be produced at STP from the reaction in part c.

Part(a) Step 1:Given information

We have been given,

The line-angle formula for ethylcyclopentane. 

Part(a) Step 2: Explanation

We know that the molecular formula of  the  ethylcyclopentane is $C_7{H}_{14}$,

Part(b) Step 1:Given information

We have to find out the balanced chemical equation for the complete combustion of ethylcyclopentane. 

Part(b) Step 2:Explanation

Now, we write the balanced chemical equation for the complete combustion of ethylcyclopentane. 

$2C_7{H}_{14}+21O_2\to 14C{O}_2+14H_{2}O$.

Part(c) Step 1:Given information

We have to find out the how many grams of $O_2$required for the combustion of $25.0g$of  ethylcyclopentane.

Part(c) Step 2: Explanation

Now, we solve this part of this question to get answer,

$2C_7{H}_{14}+21O_2\to 14C{O}_2+14H_{2}O$,

$\Rightarrow 25g=\frac{25}{98}mol$,

$\therefore 2mol C_7{H}_{14}$$\to 21 mole O_2$,

$\Rightarrow 1mol\to \frac{21}{2}$,

$\Rightarrow \frac{25}{98}mol\to \frac{25}{98}\times \frac{21}{2}mol O_2=\frac{25}{98}\times \frac{21}{2}\times 32g$,

$\Rightarrow 85.71g$

Part(d) Step 1:Given information

We have to find out how many liters of $C{O}_2$ would be produced at STP from the reaction in part c.

Part(d) Step 2:Explanation

Now, we solve this part to get answer,

$2C_7{H}_{14}+21O_2\to 14C{O}_2+14H_{2}O$,

$\Rightarrow 2mol C_7{H}_{14}\to 14 mol C{O}_{2,}$

$\Rightarrow \frac{25}{98}mol\to \frac{14}{2}\times \frac{25}{98}mol C{O}_2,$

$\Rightarrow 7\times \frac{25}{98}\times 22.4L C{O}_2=40L$