91Ó°ÊÓ

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

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

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

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 :${\mathrm{C}}_2{\mathrm{H}}_2+\frac{5}{2}{\mathrm{O}}_2→2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

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

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

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

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

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

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

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

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

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 ${\mathrm{C}}_2{\mathrm{H}}_2+\frac{5}{2}{\mathrm{O}}_2→2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

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

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