Q. 11.71 - An Introduction to General, Organic, and Biological Chemistry

Question

Consider the compound propane. $(7.4,7.7,11.2,11.4)$

a. Draw the condensed structural formula for propane.

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

c. How many grams of $O_2$ are needed to react with $12.0L$of propane gas at STP?

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

Step-by-Step Solution

Answer

(a) The condensed structural formula for propane is given below,

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

$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$.

(c) $85.71g$of $O_2$are needed to react with $12.0L$of propane gas at STP.

(d) $70.71g$of $C{O}_2$would be produced from the reaction in part c.

Part(a) Step 1: Given information

We have been given,

The condensed structural formula for propane. 

Part(a) Step 2:Explanation

Condensed Structural formula is a means of expressing molecules entirely in language. For basic hydrocarbon like propane has the condensed structural formula is $C{H}_{3}C{H}_{2}C{H}_3$,

Part(b) Step 2: Given information

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

Part(b) Step 2: Explanation

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

Incomplete combustion happens when there is insufficient oxygen for complete burning. Incomplete combustion produces water vapor, carbon dioxide, and heat once more. However, it also emits carbon monoxide.

$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$.

Part(c) Step 1:Given information

We have to find out the how many grams of $O_2$ are needed to react with $12.0L$ of propane gas at STP.

Part(c) Step 2: Explanation

Now, we solve this question to get answer,

$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$,

$12L=\frac{12}{22.4}mol$,

$1mole$of Propane$\to 5mole O_2$

$$\begin{gathered} \Rightarrow \frac{12}{22.4}mol=\frac{12}{22.4}\times 5mol, \\ \Rightarrow \frac{12}{22.4}\times 5\times 32 g=85.71 g \end{gathered}$$

Part(d) Step 1: Given information

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

Part(d) Step 2: Explanation

Now, we solve this part to get answer,

$1mole$propane$\to 3mole C{O}_2$,

$\Rightarrow \frac{12}{22.4}mole=\frac{3\times 12}{22.4}mole C{O}_2$,

$\Rightarrow \frac{3\times 12}{22.4}\times 44 grams=70.71g$.