Q. 11.67 - An Introduction to General, Organic, and Biological Chemistry

Question

The density of pentane, a component of gasoline, is $0.63 g/mL$. The heat of combustion for pentane is $845 Kcal/mole.$ $(7.2,7.4,7.7,7.8,8.6,11.2,11.4)$

a. Write the balanced chemical equation for the complete combustion of pentane.

b. What is the molar mass of pentane?

c. How much heat is produced when $1$$gal$ of pentane is burned $(1 gal=4 qt)?$

d. How many liters of $C{O}_2$ at STP are produced from the complete combustion of $1 gal$ of pentane?

Step-by-Step Solution

Answer

(a) The balanced chemical equation for the complete combustion of pentane. 

$C_2{H}_{12}+8O_2\to 5C{O}_2+6H_{2}O$.

(b) The molar mass of pentane is $72.15 g/mol$.

(c) The heat is produced when $1gal$ of pentane is burned is $27,900.94 Kcal$.

(d) $3697.12 L$of $C{O}_2$ at STP are produced from the complete combustion of $1gal$ of pentane.

Part(a) Step 1: Given information

We have been given,

The balanced chemical equation for the complete combustion of pentane. 

Part(a) Step 2: Explanation

The balanced equation for when pentane gas $C_5{H}_{12}$combusts with oxygen gas$O_2$ to form water

$H_{2}O$ and carbon dioxide $C{O}_2$is $C_2{H}_{12}+8O_2\to 5C{O}_2+6H_{2}O$.

Part(b) Step 1: Given information

We have to find out the molar mass of pentane.

Part(b) Step 2: Explanation

Because pentane is the fifth of the first ten members of the alkane family, its molar mass may be calculated simply by looking at its chemical formula, $C_5{H}_{12}$

Thus we have $5$Carbon, and the atomic number of carbon is $12$, and we have$5$, so it'll be $5C*12= 60.0g.$

When we consider Hydrogen, which has an atomic number of $1$, and we have $12H$ on hand, we get $12H *1= 12.0g.$

The molar mass of pentane is$60.0+12.0=72g$ when the two are added together.

Part(c) Step 1: Given information

We have to find out the heat is produced when $1gal$of pentane is burned.

$1gal=4qt$,

Part(c) Step 2: Explanation

Now, we solve this question to get answer,

$C_2{H}_{12}+8O_2\to 5C{O}_2+6H_{2}O$,

$\Rightarrow 1 gal=4qt=4\times \frac{1}{1.06}l=\frac{4\times 1000}{1.06}ml$,

$\therefore$total weight pentane$=\frac{0.63\times 4000}{1.06}q ms$,

$\Rightarrow \frac{4\times 630}{1.06}\times \frac{1}{72}mols$,

$\Rightarrow 33.01 moles$,

$1 mole=845Kcal$,

$\Rightarrow 33.01mole=845\times 33.01Kcal=27900.94Kcal$.

Part(d) Step 1: Given information

We have to find out how many liters of $C{O}_2$at STP are produced from the complete combustion of $1 gal$of pentane.

Part(d) Step 2: Explanation

$1 mole$of $C_5{H}_{12}$$\to 5moles$of$C{O}_2$,

$\therefore$$1 gal i.e 33.01 mol\to 5\times 33.01 moles$,

$\Rightarrow 165.05 moles$,

$\Rightarrow 165.05\times 22.4 L (at STP)$,

$\Rightarrow 3697.12L$.