Question #53a00

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Question #53a00

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Answer 1 · 2014-12-03T20:51:21

One would need $3498.9$ $3498.9$ grams of $O_{2}$ $O_2$ for this particular combustion.

The combustion reaction can be written as follows

$C_{9} H_{20} + 14 O_{2} \to 9 C O_{2} + 10 H_{2} O$ $C_9H_20 + 14O_2 -> 9CO_2 + 10H_2O$

we can see that $1$ $1$ mole of nonane needs $14$ $14$ moles of $O_{2}$ $O_2$ for the combustion reaction; from $m_{C_{9} H_{20}}$ $m_(C_9H_20)$ = $1$ $1$ kg, one can solve for the number of moles of nonane

$n_{C_{9} H_{20}} = \frac{1000 g}{128 \frac{g}{m o l}} = 7.81$ $n_(C_9H_20) = (1000g)/(128 g/(mol)) = 7.81$ moles

(using $1 k g$ $1 kg$ = $1000 g r a m s$ $1000 grams$ and knowing the molar mass of nonane to be $128 \frac{g}{m o l}$ $128g/(mol)$ );

Therefore, the number of $O_{2}$ $O_2$ moles will be

$n_{O_{2}} = 14 \cdot 7.81 = 109.34$ $n_(O_2) = 14 * 7.81 = 109.34$ moles

The mass of $O_{2}$ $O_2$ yields

$m_{O_{2}} = 109.34 \cdot 32 = 3498.9$ $m_(O_2) = 109.34 * 32 = 3498.9$ grams

IF you need the quantity of $A I R$ $AIR$ , just remember that air contains $20.95 % O_{2}$ $20.95% O_2$ , $78.09 % N$ $78.09% N$ , and a little under $1 %$ $1%$ other gasses ( $A r, C O_{2}$ $Ar, CO_2$ ,and others).

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Question #53a00

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