<u>Answer:</u> The moles of ethylene gas that can react is 0.212 moles
<u>Explanation:</u>
To calculate the moles of oxygen gas, we use the equation given by ideal gas which follows:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
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}](https://tex.z-dn.net/?f=0.0821%5Ctext%7B%20L.%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D)
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](https://tex.z-dn.net/?f=1.2atm%5Ctimes%2012.9L%3Dn%5Ctimes%200.0821%5Ctext%7B%20L.%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D%5Ctimes%20297K%5C%5C%5C%5Cn%3D%5Cfrac%7B1.2%5Ctimes%2012.9%7D%7B0.0821%5Ctimes%20297%7D%3D0.635mol)
For the given chemical equation:
![C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)](https://tex.z-dn.net/?f=C_2H_4%28g%29%2B3O_2%28g%29%5Crightarrow%202CO_2%28g%29%2B2H_2O%28g%29)
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 =
of ethylene gas
Hence, the moles of ethylene gas that can react is 0.212 moles