Question

The density of pentane, a component of gasoline, is $0.63g/mL$. The heat of combustion for pentane is $845Kcal/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 $(1gal=4qt)?$

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

Short 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 $72.15g/mol$.

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

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

Step-by-step solution

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,

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

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

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

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

$\Rightarrow 33.01moles$,

$1mole=845Kcal$,

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

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$1gal$of pentane.

Part(d) Step 2: Explanation

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

$\therefore$$1gali.e33.01mol\to 5\times 33.01moles$,

$\Rightarrow 165.05moles$,

$\Rightarrow 165.05\times 22.4L(atSTP)$,

$\Rightarrow 3697.12L$.