Question

The chemical equation for the combustion of acetylene (C2H2) is given below.

mc021-1.jpg

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

Answer 1

The balanced reaction would be:

C2H2 + 5/2O2 = 2CO2 + H2O

We are given the amount of acetylene in the reaction. This will be the starting point of our calculation. We use the ideal gas equation to find for the number of moles.

n = PV / RT = 1.00(.1005 L) / (0.08206 atm L/mol K ) 273.15 K
n= 4.4837 x 10^-3 mol C2H2

4.4837 x 10^-3 mol C2H2 (5/2 mol O2/ 1 mol C2H2) = 0.0112 mol O2

V = nRT/P = 0.0112 mol O2 x 273.15 K x 0.08206 atm L/mol K / 1 atm
V=0.25125 L or 251.25 mL

Answer 2

251.25 mL of O₂

Solution:

The balance chemical equation is as follow,

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

As we know if the gas is acting ideally then 1 mole of any gas at standard temperature and pressure it will occupy exactly 22.4 L or 22400 mL of volume.

Keeping this in mind according to equation,

44800 mL (2 mol) of C₂H₂ required = 112000 mL (5 mol) of O₂

So,

100.50 mL of C₂H₂ will require = X mL of O₂

Solving for X,

X = (100.5 mL × 112000 mL) ÷ 44800 mL

X = 251.25 mL of O₂