Question

Picture *** pls help

Answer 1

The volume of Oxygen at STP required : 251.25 L

Further explanation

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.

Reaction(combustion) :

2C₂H₂(g)+5O₂(g)⇒4CO₂(g)+2H₂O(g)

mol C₂H₂ :

$\tt \dfrac{100.5}{22.4}=4.49$

mol ratio C₂H₂ : O₂ = 2 : 5, so mol O₂ :

$\tt \dfrac{5}{2}\times 4.49=11.23$

volume  O₂ at STP :

$\tt 11.23\times 22.4=251.55~L$

Or you can use Avogadro's hypothesis :

In the same temperature and pressure, in the same volume conditions, the gas contains the same number of molecules  

So the ratio of gas volume will be equal to the ratio of gas moles  :

$\tt \dfrac{V_1}{V_2}=\dfrac{n_1}{n_2}\\\\\dfrac{100.5}{V_2}=\dfrac{2}{5}\\\\V_2=\dfrac{100.5\times 5}{2}=251.25~L$