How many liters of hydrogen gas are required to react with 3.5" liters of oxygen gas in the following reaction? 2H2(g)+O2(g) -->2H2O(g)

7 L of H₂.

Explanation:

The balanced equation for the reaction is given below:

2H₂ + O₂ —> 2H₂O

From the balanced equation above,

1 L of O₂ required 2 L of H₂.

Finally, we shall determine the volume of H₂ required to react with 3.5 L of O₂. This can be obtained as follow:

From the balanced equation above,

1 L of O₂ required 2 L of H₂.

Therefore, 3.5 L of O₂ will require

= 3.5 × 2 = 7 L of H₂.

Thus, 7 L of H₂ is required to for the reaction.

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.