Answer:
From 2 liters of hydrogen gas and 2 liters of nitrogen gas, we can produce 1.10 L of ammonia.
Step-by-step explanation:
First of all, think the ballanced equation:
3H₂ + N₂ → 2NH₃
We must know the density of products and reactants to solve this.
Density shows the relation between mass and volume of a compound, like this:
ρ = mass /volume
ρ H₂ = 0.071 g/L
ρ N₂ = 1.2506 g/L
ρ NH₃ = 0.73 g/L
Let's calculate the mass of reactants
mass of H₂ = ρ H₂ . volume H₂
mass of H₂ = 0.071 g/L . 2L → 0.142 g
mass of N₂ = ρ N₂ . volume N₂
mass of N₂ = 1.2506 g/L . 2L → 2.50 g
Now, let's calculate the moles
mass H₂ / molar mass H₂ = moles
0.142 g / 2g/m = 0.071 moles
mass N₂ / molar mass N₂ = moles
2.50 g /28 g/m = 0.089 moles
As the ratio between products is 3:1, it's easy to see that the limiting reactant is H₂. Let's see
1 mol of N₂ needs to react with 3 moles of H₂
0.089 mol of N₂ need to react with (0.089 .3) /1 = 0.267
I only have 0.071 moles.
The ratio between H₂ and NH₃ is 3:2 so the rule of three is:
3 moles of H₂ __ are necessary to make 2 moles of ammonia
0.071 moles of H₂ _ are necessary to make ( 0.071 .2)/ 3 = 0.0473 moles
Let's convert this moles in mass, with the molar mass
Moles . Molar mass = mass
0.0473 moles . 17 g/m = 0.804 g
Now, with the density we can know the volume:
ρ NH₃ = 0.73 g/L
Volume NH₃ = mass NH₃ /ρ NH₃
Volume NH₃ = 0.804 g/0.73 g/L
Volume NH₃ = 1.10L