Question

Help please!!!

A chemical reaction produces oxygen gas. You collect 3.00×1023 oxygen gas molecules at STP .


What volume of oxygen gas did you collect?

Answer 1

We will collect 11.17 liters of oxygen gas.

Step-by-step explanation:

`N=n* N_A`

Where:

N = Number of particles / atoms/ molecules

n = Number of moles

`N_A=6.022* 10^(23) mol^(-1)` = Avogadro's number

We have:

Molecules of oxygen gas N =
`3.00* 10^(23)`

n =?

`n=(N)/(N_A)=(3.00* 10^(23))/(6.022* 10^(23) mol^(-1))`

n = 0.4982 moles of oxygen

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 1 atm (at STP)

V = Volume of gas = 10 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 273.15 K (at STP)

Putting values in above equation, we get:

`V=(nRT)/(P)=(0.4982 mol* 0.0821 atm L/mol K* 273.15 K)/(1 atm)`

V = 11.17 L

We will collect 11.17 liters of oxygen gas.