Ogen gas and nitrogen gas react to produce ammonia. how many liters of ammonia can be produced from 2 liters of hydrogen gas and 2 liters of nitrogen gas at stp

From 2 liters of hydrogen gas and 2 liters of nitrogen gas, we can produce 1.10 L of ammonia.

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

screening

explanation:

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"