Question #52b92

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Question #52b92

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Answer 1 · 2014-03-16T22:36:54

The balance chemical equation is
2 $C_{2}$ $C_2$ $H_{6}$ $H_6$ + 7 $O_{2}$ $O_2$ ---> 4C $O_{2}$ $O_2$ + 6 $H_{2}$ $H_2$ O

as per the equation : 2 moles of $C_{2}$ $C_2$ $H_{6}$ $H_6$ needs 7 moles of $O_{2}$ $O_2$ .

moles of $C_{2}$ $C_2$ $H_{6}$ $H_6$ = Volume of $C_{2}$ $C_2$ $H_{6}$ $H_6$ / 22.4 L

moles of $C_{2}$ $C_2$ $H_{6}$ $H_6$ = 16.4 L / 22.4 L = 0.73 mol

as per the molar ratio X mol of $C_{2}$ $C_2$ $H_{6}$ $H_6$ will need react with 0.98 mol of $O_{2}$ $O_2$

2 mol of $C_{2}$ $C_2$ $H_{6}$ $H_6$ /7 mol of $O_{2}$ $O_2$ =
X mol of $C_{2}$ $C_2$ $H_{6}$ $H_6$ / 0.98 mol of $O_{2}$ $O_2$

7.x = 0.98 x 2
7x = 1.96 , x = 1.96/ 7 = 0.28 mol
0.28 moles of $C_{2}$ $C_2$ $H_{6}$ $H_6$ can react with 0.98 mol of $O_{2}$ $O_2$ .
All of the oxygen will be used to react with 0.28 mol of $C_{2}$ $C_2$ $H_{6}$ $H_6$ hence it is a limiting reagent. 0.73 - 0.28 = 0.45 moles of $C_{2}$ $C_2$ $H_{6}$ $H_6$ will remain unused , so it is an excess reagent.

Unused mass of $C_{2}$ $C_2$ $H_{6}$ $H_6$ =
Unused moles of $C_{2}$ $C_2$ $H_{6}$ $H_6$ x molar mass of $C_{2}$ $C_2$ $H_{6}$ $H_6$
= 0.45 mol x 30 g $m o l^{- 1}$ $mol^(-1)$ = 13.5 g.

As for the volume of $C O_{2}$ $CO_2$ produced, we know from the balanced equation that $2$ $2$ moles of $C_{2} H_{6}$ $C_2H_6$ will produce $4$ $4$ moles of $C O_{2}$ $CO_2$ ; thus, the moles of $C O_{2}$ $CO_2$ produced will be

$n_{C O_{2}} = 0.28 \cdot 2 = 0.56$ $n_(CO_2) = 0.28 * 2 = 0.56$ moles

Therefore, from $n = \frac{V}{V_{m o l}}$ $n = V/V_(mol)$ , we have

$V = 22.4 \cdot 0.56 = 12.54 L$ $V = 22.4 * 0.56 = 12.54 L$

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Question #52b92

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