Question

for complete oxidation of 2 lt of NO(g) to NO2(g) at STP ,the volume of O2 required at STP is
1. 0.5L
2. 2L
3. 1L
4. 1.5L​

Answer 1

Answer: 3. 1L

Explanation:-

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the gas = 1 atm  

V= Volume of the gas = 2 L

T= Temperature of the gas = 0°C = 273 K  

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= ?

$n=\frac{PV}{RT}=\frac{1\times 2}{0.0821\times 273}=0.09moles$

According to stoichiometry :

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

2 moles of $NO$ requires 1 mole of oxygen  

Thus 0.09 moles of $NO$ require=$\frac{1}{2}\times 0.09=0.045$ moles of oxygen  

Volume of oxygen $moles\times {\text {molar volume}}=0.045\times 22.4L=1L$

Thus the volume of $O_2$ required at STP is 1 L.