Ammonia, at stp, 1.5L of N2 reacts with 4.5L of H2, if N2 and H2 are consumed, what volume of NH3 at stp will be produced?
N2 + 3H2 ==> 2NH3
When using equations will all gases, one may use L as if they were moles.
1.5L N2 will produce 1.5 x (2/1) = 3.0L NH3 and
4.5L H2 will produce 4.5 x (2/3) = 3.0 L NH3.
So the answers is that 3.0 L NH3 will be produced at STP.
gyiofucout
To find the volume of NH3 produced, we need to determine the stoichiometry of the reaction and use it to calculate the volume.
The balanced equation for the reaction is:
N2 + 3H2 → 2NH3
From the balanced equation, we can see that for every 1 mole of N2, 2 moles of NH3 are produced. We can also see that for every 3 moles of H2, 2 moles of NH3 are produced. This means that the reaction consumes N2 and H2 in a 1:3 ratio.
Let's calculate the moles of N2 and H2 available for the reaction:
Moles of N2 = Volume of N2 / Molar volume at STP
Molar volume at STP is 22.4 L/mol
Moles of N2 = 1.5 L / 22.4 L/mol ≈ 0.067 moles
Moles of H2 = Volume of H2 / Molar volume at STP
Moles of H2 = 4.5 L / 22.4 L/mol ≈ 0.201 moles
Since the reaction consumes N2 and H2 in a 1:3 ratio, we can determine which reactant is limiting by comparing the moles. In this case, N2 is the limiting reactant because it has the smaller number of moles.
To find the moles of NH3 produced, we use the stoichiometry of the reaction. Since 1 mole of N2 produces 2 moles of NH3, we can calculate:
Moles of NH3 = Moles of N2 × (2 moles of NH3 / 1 mole of N2)
Moles of NH3 = 0.067 moles × (2 moles of NH3 / 1 mole of N2) ≈ 0.134 moles
Now, let's calculate the volume of NH3 produced:
Volume of NH3 = Moles of NH3 × Molar volume at STP
Volume of NH3 = 0.134 moles × 22.4 L/mol ≈ 3.0 L
Therefore, at STP, approximately 3.0 liters of NH3 will be produced.