The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>
The equation for the complete combustion of C2H2 is as below:

The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
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The equilibrium constant k is actually the ratio of the
concentration of the products over the concentration of reactants at equilibrium. So if the
concentration of products < concentration of reactants, therefore the
constant k will be small. But if the concentration of products >
concentration of reactants, the constant k will be large. In this case the
value is too small (x10^-19), therefore we can say that the reaction favors the
reactant side:
the equilibrium lies far to the left
Answer:
100 C or 212 F
Explanation:
That is the boiling point for liquid so both gas and liquid are presented.
It can hold 4 minutes more electrons.