A 0.228 mol sample of Ar gas is contained in a 6.00 L flask at room temperature and pressure. What is the density of the gas, in grams/liter, under these conditions?

Density of Ar gas under the given condition is 1.52 g/L

Explanation:

Density in grams/liter = (mass of Ar gas present in flask)/(volume of flask)

We know, number of moles = (mass)/(molar mass)

Molar mass of Ar = 39.948 g/mol

So, mass of 0.228 mol of Ar =

Volume of flask = 6.00 L

So, density of gas inside flask =

Hence, density of Ar gas under the given condition is 1.52 g/L

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.