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

the answer is true.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.