What is the volume occupied by 40 grams of argon gas (Ar) at standard conditions?

volume occupied by 40 grams of argon gas (Ar) at standard conditions = 22.443L

Explanation:

Using the ideal gas law that

PV =nRT  

Where p  =pressure,  

V = volume,  

n = number of moles  

R = gas constant, and  

T= temperature in Kelvin.

Since Argon is at STP( Standard conditions), This means that temperature is   273.15 K  and pressure is  1 atm

Also the  gas constant= 0.08206 L atm/K mol . we  are using this value because its  units match the units of our values to aid easy calculation

So,

Number of moles = Mass/ Molar mass

=40g/39.948 g/mol

Number of moles of Argon=1.0013moles.

Now Bringing

pV= nRT

V= nRT/p

=(1.0013mol x 0.08206 L atm/K mol x 273.15 K  ) /1atm

Volume =22.443L

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

i believe it is a.) released into

explanation: