Question

The no of liters of air required to burn 8 liters of c2h2 approximately is
Pls answer fast

Answer 1

Answer-

We need a stoichiometric equation,.......and get an approx. Volume of 100.L air.....

Explanation-

We need a stoichiometric equation to represent the combustion:

$HC=CH(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)$

And thus given that at constant pressure, and constant temperature (conditions that must be reasonably assumed), one volume of acetylene gas required $\frac{5}{2}$ volumes of dioxygen for complete combustion....

And so, if there are 8. L to acetylene, we requires a volume

of $8.L\times \frac{5}{2}=??L$ dioxygen gas.....

But we are not finished there (says he, finally reading the question). The percentage of oxygen in air BY volume is approx 21%.

And thus we need a volume of....

$\frac{20.L.O_2}{O.21.L.O_2.L^{-1} air}$ ≅$100.L.air..$

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Answer 2

Answer:

The correct answer to this question is the oxygen content of air, measured by volume, is around 21%.

Explanation:

Given - The no of liters of air required to burn 8 liters of $c_{2}h_{2}$

To Find - The no of liters of air required to burn 8 liters of c2h2 approximately is

An equation in stoichiometry to depict combustion.

$HC = CH + \frac{5}{2}O_{2}$  → $2CO_{2}(g) +H_{2}O$

Given that one volume of acetylene gas requires quantities of dioxygen for full combustion at constant pressure and constant temperature (conditions that must be logically established),

To put it another way, if there are 8. L of acetylene, we need a volume.

The oxygen content of air, measured by volume, is around 21%.

$100.L.air$

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