An unknown gas effuses at a speed one quarter of that of helium. what is the molar mass of the unknown gas? it is either sulfur di oxide or sulfur tri oxide. which gas is it.

sulfur dioxide.

Explanation:

Rate of effusion ∝ 1/√molar mass.

(Rate of effusion of He) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of He).

∵ Rate of effusion of unknown gas = 1/4 (Rate of effusion of He)

∴ (Rate of effusion of He) / (Rate of effusion of unknown gas) = 4.

Molar mass of He = 4.0 g/mol.

∵ (Rate of effusion of He) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of He).

∴  4.0 = (√molar mass of unknown gas) / √4 = (√molar mass of unknown gas) / 2.

∴ (√molar mass of unknown gas) = 8.0.

By squaring the both sides:

∴ molar mass of unknown gas = 64.0 g/mol.

The molar mass of sulfur dioxide = 64.0 g/mol and the molar mass of sulfur trioxide = 80.0 g/mol.

So, the unknown gas is sulfur dioxide.

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.

where is the following

explanation: