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

Answer:

We will collect 11.17 liters of oxygen gas.

Explanation:

$N=n\times N_A$

Where:

N = Number of particles / atoms/ molecules

n = Number of moles

$N_A=6.022\times 10^{23} mol^{-1}$ = Avogadro's number

We have:

Molecules of oxygen gas N = $3.00\times 10^{23}$

n =?

$n=\frac{N}{N_A}=\frac{3.00\times 10^{23}}{6.022\times 10^{23} mol^{-1}}$

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:

$V=\frac{nRT}{P}=\frac{0.4982 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}$

V = 11.17 L

We will collect 11.17 liters of oxygen gas.