Question 1 a sample of 0.255 mole of gas has a volume of 748 ml at 28°c. calculate the pressure of this gas. (r= 0.0821 l ∙ atm / mol ∙ k) 0.784 atm 8.42 atm 0.00842 atm 7.84 × 10-4 atm none of the above

Answer : The pressure of the gas is, 8.42 atm

Explanation :

Using ideal gas equation,

where,

P = pressure of the gas = ?

V = volume of the gas = 748 ml = 0.748 L

conversion used : (1 L = 1000 ml)

T = temperature of the gas =

n = number of moles of the gas = 0.255 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get  the pressure of the gas.

Therefore, the pressure of the gas is, 8.42 atm

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.