Q. 7.115 - An Introduction to General, Organic, and Biological Chemistry

Question

Acetylene gas, ${\mathrm{C}}_2{\mathrm{H}}_2$, used in welders' torches, releases $1300 \mathrm{kJ}$ of heat when one mole of ${\mathrm{C}}_2{\mathrm{H}}_2$ undergoes combustion.
a. Write a balanced equation for the reaction, including the heat of reaction.
b. Is the reaction endothermic or exothermic?

c. How many moles of ${\mathrm{H}}_2\mathrm{O}$ are produced when  $2.00$ moles of ${\mathrm{O}}_2$ reacts?
d. How many grams of ${\mathrm{O}}_2$ are needed to react with $9.80 \mathrm{g}$ of ${\mathrm{C}}_2{\mathrm{H}}_2$

Step-by-Step Solution

Answer

Part(a) The balanced equation is ${\mathrm{C}}_2{\mathrm{H}}_2+\frac{5}{2}{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

Part(b) The reaction is exothermic.

Part(c) $0.8$ moles of ${\mathrm{H}}_2\mathrm{O}$ are produced when $2.00$ moles of ${\mathrm{O}}_2$ reacts.

Part(d) $15.2 \mathrm{g} {\mathrm{O}}_2$ is needed to react with $9.8 \mathrm{g} {\mathrm{C}}_2{\mathrm{H}}_2$

Part(a) Step 1 : Given information

We are given that Acetylene gas, is used in welders' torches, releases $1300 \mathrm{kJ}$ of heat when one mole of acetylene undergoes combustion.

Part(a) Step 2 : Simplify

As given in the question acetylene is used in welder's torches and release energy during combustion. Therefore unbalanced equation will be : 

${\mathrm{C}}_2{\mathrm{H}}_2+{\mathrm{O}}_2\to {\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

Now in equation on left side there are two carbon , hydrogen and oxygen are present whereas in right side one carbon two hydrogen and three oxygen are present. so to balance the equation : ${\mathrm{C}}_2{\mathrm{H}}_2+\frac{5}{2}{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

Part(b) Step 1 : Given information

We are given that Acetylene gas, is used in welders' torches, releases $1300 \mathrm{kJ}$ of heat when one mole of acetylene undergoes combustion.

Part(b) Step 2 : Simplify

As given in the question during combustion energy is released, Therefore the reaction used in welder's torches is Exothermic.

Part(c) Step 1 : Given information

We are given that Acetylene gas, is used in welders' torches, releases $1300 \mathrm{kJ}$ of heat when one mole of acetylene undergoes combustion.

Part(c) Step 2 : Simplify

The reaction is ${\mathrm{C}}_2{\mathrm{H}}_2+\frac{5}{2}{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

For two moles of oxygen, multiply the equation with $\frac{4}{5}$

After multiplying , we get $\frac{4}{5}{\mathrm{C}}_2{\mathrm{H}}_2+2{\mathrm{O}}_2\to \frac{8}{5}{\mathrm{CO}}_2+\frac{4}{5}{\mathrm{H}}_2\mathrm{O}$

So from above equation, We can say $0.8$ moles of water is produced.

Part(d) Step 1 : Given information

We are given that Acetylene gas, is used in welders' torches, releases $1300 \mathrm{kJ}$ of heat when one mole of acetylene undergoes combustion.

Part(d) Step 2 : Simplify

As we know, Molar mass of acetylene is $26.0 \mathrm{g}$

so $9.80 \mathrm{g}$means $0.38$ moles of acetylene. According to reaction ${\mathrm{C}}_2{\mathrm{H}}_2+\frac{5}{2}{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

$$\begin{gathered} 1 \mathrm{mole} {\mathrm{C}}_2{\mathrm{H}}_2=2.5 \mathrm{mole} {\mathrm{O}}_2 \\ 0.38 \mathrm{mole} {\mathrm{C}}_2{\mathrm{H}}_2=0.95 \mathrm{mole} {\mathrm{O}}_2 \\ 1 \mathrm{mole} {\mathrm{O}}_2=16 \mathrm{g} \\ 0.95 \mathrm{mole} {\mathrm{O}}_2=15.2 \mathrm{g} \end{gathered}$$

$15.2 \mathrm{g} {\mathrm{O}}_2$ is needed to react.