# Question #52b92

Mar 16, 2014

The balance chemical equation is
2 ${C}_{2}$${H}_{6}$ + 7${O}_{2}$ ---> 4C${O}_{2}$ + 6${H}_{2}$O

as per the equation : 2 moles of ${C}_{2}$${H}_{6}$ needs 7 moles of ${O}_{2}$ .

moles of ${C}_{2}$${H}_{6}$ = Volume of ${C}_{2}$${H}_{6}$ / 22.4 L

moles of ${C}_{2}$${H}_{6}$ = 16.4 L / 22.4 L = 0.73 mol

as per the molar ratio X mol of ${C}_{2}$${H}_{6}$ will need react with 0.98 mol of ${O}_{2}$

2 mol of ${C}_{2}$${H}_{6}$ /7 mol of ${O}_{2}$ =
X mol of ${C}_{2}$${H}_{6}$ / 0.98 mol of ${O}_{2}$

7.x = 0.98 x 2
7x = 1.96 , x = 1.96/ 7 = 0.28 mol
0.28 moles of ${C}_{2}$${H}_{6}$ can react with 0.98 mol of ${O}_{2}$.
All of the oxygen will be used to react with 0.28 mol of ${C}_{2}$${H}_{6}$ hence it is a limiting reagent. 0.73 - 0.28 = 0.45 moles of ${C}_{2}$${H}_{6}$ will remain unused , so it is an excess reagent.

Unused mass of ${C}_{2}$${H}_{6}$ =
Unused moles of ${C}_{2}$${H}_{6}$ x molar mass of ${C}_{2}$${H}_{6}$
= 0.45 mol x 30 g $m o {l}^{- 1}$ = 13.5 g.

As for the volume of $C {O}_{2}$ produced, we know from the balanced equation that $2$ moles of ${C}_{2} {H}_{6}$ will produce $4$ moles of $C {O}_{2}$; thus, the moles of $C {O}_{2}$ produced will be

${n}_{C {O}_{2}} = 0.28 \cdot 2 = 0.56$ moles

Therefore, from $n = \frac{V}{V} _ \left(m o l\right)$, we have

$V = 22.4 \cdot 0.56 = 12.54 L$