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What volume of oxygen at STP is required for the complete combustion of 100.50 mL of C2H2?

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The given substance combusts following the reaction:

C2H2 + (5/2)O2 -> 2CO2 + H2O

Assume C2H2 is an ideal gas. At STP, 1 mol of an ideal gas
occupies 22.4 L. Given 100.50 mL of C2H2, this means that there is 4.4866 x
10^(-3) mol. Combusting 1 mol of C2H2 consumes (5/2) mol of O2, then combusting
the given amount of C2H2 consumes 0.01121 mol of O2. At STP, this amount of O2
occupies 251.25 mL. 

Answer: 251.25 ml

Explanation: \(2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2+2H_2O\)

According to Avogadro's law, 1 mole of every gas occupies 22.4 L at Standard temperature and pressure (STP).

2 moles of  \(C_2H_2(g)\) occupy = \(2\times 22.4L=44.8L=44,800ml\)

5 moles of \(O_2(g)\) occupy = \(5\times 22.4L=112L=112000ml\)

Thus 44800 ml of \(C_2H_2(g)\) reacts with 112000 ml of \(O_2(g)\) at STP

100.50 ml of \(C_2H_2(g)\) reacts with =\(\frac{112000}{44800}\times 100.50=251.25 ml\) of \(O_2(g)\) at STP.


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