Answer :
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: [tex]2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2+2H_2O[/tex]
According to Avogadro's law, 1 mole of every gas occupies 22.4 L at Standard temperature and pressure (STP).
2 moles of [tex]C_2H_2(g)[/tex] occupy = [tex]2\times 22.4L=44.8L=44,800ml[/tex]
5 moles of [tex]O_2(g)[/tex] occupy = [tex]5\times 22.4L=112L=112000ml[/tex]
Thus 44800 ml of [tex]C_2H_2(g)[/tex] reacts with 112000 ml of [tex]O_2(g)[/tex] at STP
100.50 ml of [tex]C_2H_2(g)[/tex] reacts with =[tex]\frac{112000}{44800}\times 100.50=251.25 ml[/tex] of [tex]O_2(g)[/tex] at STP.