If 1.4 L of nitrogen reacted, calculate the volume of hydrogen gas needed to react. Show your work.

Question

ogen gas reacts with hydrogen gas to form ammonia according to the equation: 

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Answer 1

The chemical reaction equation is already balanced. So, according to this balanced equation, you need 3 moles of Hydrogen gas for every 1 mole of Nitrogen gas. Now we need to find how many moles of Nitrogen is there in 1.4 L of nitrogen. If you remember in the chapter discussing gases and gas laws, there is a conversion factor which is .... At standard conditions (STP), 1 mole of ANY gas would occupy 22.4 L of volume. In other words, if you have 22.4 L of nitrogen, it means you have 1 mole of nitrogen [Remember, mole is just a number of items, but HUGE number => 6.023 x 10²³ items in 1 mole]

Anyway,

22.4 L = 1 mole

Therefore 1.4 L of N₂ ==> (1 mol N₂/22.4 L N₂) * (1.4 L N₂) = 0.0625 mol of N₂

According to the equation, each mole of N₂ needs 3 moles of H₂

Therefore, 0.0625 mole of N₂ would need => (3 mol H₂/1 mol N₂)*(0.0625 mol N₂) => 0.1875 mol of H₂

Now you need to convert the moles of H₂ to Liters of H₂ by using the same conversion factor.

==> (22.4 L/1 mol)*(0.1875 mol) ==> 4.2 L of H₂