Question

An amount of $2$ moles $KCl{O}_3$ is decomposed completely to produce $O_2$ gas. How many moles of butane, $C_4{H}_2$ can be burnt completely by the $O_2$ gas produced?

• A. $0.6$
• B. $1.0$
• C. $2.0$
• D. $3.0$

Answer 1

The correct option is A $0.6$

Given $2$ mole $KCl{O}_3$ is decomposed to $O_2$ gas

$2KCl{O}_3⟶2KCl+3O_2$

We know by this reaction

$2$ mole $KCl{O}_3$ produces $3$ mole $O_2$

Now, given

$C_4{H}_2$ is burnt using the above $O_2$

Reaction for burning of $C_4{H}_2$ is

$2C_4{H}_2+9O_2⟶8C{O}_2+2H_{2}O$

i.e. $2$ moles of $C_4{H}_2$ requires $9$ moles of $O_2$ for complete burning

but we are provided with $3$ moles $O_2$

So, now $O_2$ is the limiting reagent here.

$2$ moles $C_4{H}_2$ require $9$ moles $O_2$

$\Rightarrow 3$ moles $O_2$ can burn$⟶\frac{3\times 2}{9}$ moles of $C_4{H}_2$

$\therefore$ Moles of butane $C_4{H}_2$ burnt= $\frac{6}{9}$

$=0.6$