Approach: We need to get all gases into the same unit, and then we need to find the number of particles from here.
Recall:
Number of particles = (Avogadro’s Number)(number of moles)
Avogadro’s number ~ $(6.02)(10)^{23}
For this question, we only need to compare the number of moles since, as you can see, more moles = more particles. You should memorize Avogadro’s regardless because it could be needed for other questions.
Let’s start by converting everything to moles:
Number of Moles = mass / molecular mass
N = m/M
A. 33.6 L of chlorine gas
We are told that 1 mole of gas = 22.4L
\frac{33.6L}{22.4L/mol} = 1.5mol
B. 66.0 g of carbon dioxide gas
CO_2
Molecular Formula (M): 12g/mol + 2(16g/mol)
M = 44g/mol
n= \frac{66g}{44g/mol} = 1.5 mol
C. 22.4 L of hydrogen gas
\frac{22.4L}{22.4L/mol} = 1 mol
D. 10.0 g of helium gas
M = 4g/mol
n= \frac{10g}{4g/mol} = 2.5 mol
E. 64.0 g of oxygen gas
Remember the diatomic gases when determining molecular weight: HOFBrINCl
Since oxygen is diatomic, its molecular weight in the form of oxygen gas is:
M = (2)(16g/mol) = 32g/mol
n= \frac{64g}{32g/mol} = 2 mol
As we can see, Helium has the most moles and therefore it has the most particles.
\fcolorbox{red}{grey!30}{Therefore D is correct. }