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Air contain 20 % $\ce{O2}$ by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene?

(a) 1064 cm³
(b) 212.8 cm³
(c) 500 cm³
(d) 1250 cm³

Answer: (d)

My solution:

equation: $$\ce{2C2H2 +5O2 -> 4CO2 +2H2O}$$

From stoichiometric calculations:

2 $\ce{C2H2}$ moles will react with 5 molecules of oxygen
1 $\ce{C2H5 ->} 5/2$ moles of $\ce{O2}$
$22.4\ \mathrm{l}$ $\ce{C2H2 ->}$ $5/2\times22.4\ \mathrm{l}$ of $\ce{O2}$

1 volume of air -> 0.2 (20/100) volume of oxygen

Now the only thing I don’t know is how to convert the volume into liters so that I can find the correct answer.

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Let's begin with your answer:

$2\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5\ \mathrm{mol}$ of oxygen.
$1\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5/2\ \mathrm{mol}$ of $\ce{O2}$.

If we suppose that the gases are under standard conditions, then
$22.4\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 22.4\ \mathrm{L }$ $\ce{O2}$.

This means that: $0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\ \mathrm{L}$ $ce{O2}$.
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\times 5\ \mathrm{L}$ of air.
So, the volume of air is $1.25\ \mathrm{L}= 1250\ \mathrm{cm^3}$

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What I suggest is we can use Gay Lussac's law of combining volumes of gases. According to this law, 2 volumes of acetylene reacts with 5 volumes of oxygen. From the balanced equation, the ratio between volumes of gaseous reactants and products are in the ratio 2:5:4.

2 volumes of $\ce{C2H2}$ requires 5 volumes of $\ce{O2}$.

The volume of $\ce{O2}$ used for $\pu{100 cm3}$ of $\ce{C2H2} = (5/2) \times 100 = \pu{250 cm3}$.

But air contains only 20% of $\ce{O2}$, i.e. $0.2$ of $\ce{O2}$.

Thus, the amount of air required to get $\pu{250 cm3}$ of oxygen $= 250×100÷20 = \pu{1250 cm3}$

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