Question

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

Answer 1

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* 22.4L=44.8L=44,800ml`

5 moles of
`O_2(g)` occupy =
`5* 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 =
`(112000)/(44800)* 100.50=251.25 ml` of
`O_2(g)` at STP.

Answer 2

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.