Question

How many moles of ethylene (C2H4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin?

C2H4(g) + 3O2(g) yields 2CO2(g) + 2H2O(g)

Answer 1

3 moles of oxygen will react with 1 mole of ethylene. Convert 12.9 L of oxygen to x moles of oxygen, then divide by three.

Answer 2

Answer: The moles of ethylene gas that can react is 0.212 moles

Explanation:

To calculate the moles of oxygen gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 1.2 atm  

V = Volume of the gas = 12.9 L

T = Temperature of the gas = 297 K

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$1.2atm\times 12.9L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 297K\\\\n=\frac{1.2\times 12.9}{0.0821\times 297}=0.635mol$

For the given chemical equation:

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)$

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 1 mole of ethylene gas

So, 0.635 moles of oxygen gas will react with = $\frac{1}{3}\times 0.635=0.212mol$ of ethylene gas

Hence, the moles of ethylene gas that can react is 0.212 moles