Question #56a51

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Question #56a51

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Answer 1 · 2014-12-03T21:10:54

The answer is $0.667$ $0.667$ .

The combustion of acetylene can be written as

$2 C_{2} H_{2} + 5 O_{2} \to 4 C O_{2} + 2 H_{2} O$ $2C_2H_2 + 5O_2 -> 4CO_2 + 2H_2O$

The balanced chemical reaction shows a $2 : 5$ $2:5$ mole ratio for $C_{2} H_{2}$ $C_2H_2$ and $O_{2}$ $O_2$ , a $1 : 2$ $1:2$ ratio for $C_{2} H_{2}$ $C_2H_2$ and $C O_{2}$ $CO_2$ , and a $1 : 1$ $1:1$ ratio for $C_{2} H_{2}$ $C_2H_2$ and $H_{2} O$ $H_2O$ .

Now, for simplicity, let us assume that we have $1$ $1$ mol of $C_{2} H_{2}$ $C_2H_2$ to start with ( the equivalence of $26$ $26$ grams ); given the fact that the oxygen is in excess, acetylene will always act as a limiting reagent.

So, one would get $2$ $2$ moles of $C O_{2}$ $CO_2$ and $1$ $1$ mole of $H_{2} O$ $H_2O$ in the products, for a total of $n_{T} = 2 + 1 = 3$ $n_T = 2 + 1 = 3$ moles of product.

Therefore, the mole fraction of $C O_{2}$ $CO_2$ would be

$n_{C O_{2}} = (\frac{n_{C O_{2}}}{n_{T}}) = \frac{2}{3} = 0.667$ $n_(CO_2) = (n_(CO_2)/n_T) = 2/3 = 0.667$

In this case, $20 %$ $20%$ excess $O_{2}$ $O_2$ would mean that you started with $3$ $3$ moles of $O_{2}$ $O_2$ instead of $2.5$ $2.5$ , but I don't see how this would be relevant other than determining that $O_{2}$ $O_2$ is not a limiting reagent.

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Question #56a51

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