What volume is occupied by 0.34 moles of Helium gas?

I will assume that you mean

STP (Standard Temperature and Pressure)

21

3

≈

7 L

Explanation:

Since we know that

1 mole

of an ideal gas occupies

22.4 L

at

STP

then can establish the following relationships:

22.4 L

1 mole

or

1 mole

22.4 L

Knowing that, we can solve for the volume occupied by the helium gas,

0.35

mole

1

⋅

22.4 L

1

mole

=

7.84 L

as

7.84 L

−

−

−

−

−

The answer makes sense because if

1 mole

occupies

22.4 L

, then

0.35 moles

(roughly a third of the moles) will occupy a third of

22.4 L

(

if we round down 22.4 L to 21 L

)

−

−

−

21

3

≈

7 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.

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