Answer: The volume of hydrogen gas required for the given amount of ethylene gas is 113 L
Explanation:
At STP:
1 mole of a gas occupies 22.4 L of volume
We are given:
Volume of ethylene = 113 L
For the given chemical equation:
[tex]H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)[/tex]
By Stoichiometry of the reaction:
[tex](1\times 22.4)L[/tex] of ethylene reacts with [tex](1\times 22.4)L[/tex] of hydrogen gas
So, 113 L of ethylene gas will react with = [tex]\frac{(1\times 22.4)}{(1\times 22.4)}\times 113=113L[/tex] of hydrogen gas
Hence, the volume of hydrogen gas required for the given amount of ethylene gas is 113 L