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