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

Step-by-step 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* 12.9L=n* 0.0821\text{ L. atm }mol^(-1)K^(-1)* 297K\\\\n=(1.2* 12.9)/(0.0821* 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 =
`(1)/(3)* 0.635=0.212mol` of ethylene gas

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