20L of N2 and 100 L of H2 are reacted to produce NH3 what is the decrease in the V of the mixture

N2+3H2--> 2NH3
Ans me with reason

This is a limiting reagent (LR) problem. You can tell because amounts are give for both reactants.

V is voltage, value, volume. Volume maybe.
........N2 + 3H2 ==> 2NH3
begin volume = 20 + 100 = 120 L.
What volume NH3 will be produced if we use all of the N2 and and excess of H2.
Thats 20L x (2 mols NH3/1 mol N2) = 40 L.

What volume NH3 will be produced if we use all of the H2 and and excess of N2. That's 100 L x (2 mol NH3/3 mol H2) = about 67 L NH3.
You see there are two values for the amount of NH3 which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. That means N2 is the limiting reagent. We won't have any N2 left over unreacted and we will produce 40 L NH3.

How much H2 is left over? How much was used? That's 20 L N2 x (3 mol H2/1 mol N2) = 60 L H2 used. We had 100 initially so we must have 100-60 = 40 L left over. Assuming the pressure doesn't change we will have 40 L H2 which was an excess and we will have 40 L NH3 produced from the N2 so the total is????.

Answer me

To determine the decrease in the volume (V) of the mixture when 20L of N2 and 100L of H2 are reacted to produce NH3, we need to calculate the change in the number of moles of gases involved in the reaction.

According to the balanced chemical equation:
N2 + 3H2 --> 2NH3

We can see that for every 1 mole of N2 reacted, 2 moles of NH3 are produced, and for every 3 moles of H2 reacted, 2 moles of NH3 are also produced.

Given that we have 20L of N2 and 100L of H2, we can convert these volumes to moles using the ideal gas law. Assuming standard temperature and pressure (STP), 1 mole of ideal gas occupies 22.4L.

For N2:
20L * (1 mol/22.4L) = 0.89375 moles of N2

For H2:
100L * (1 mol/22.4L) = 4.46429 moles of H2

Based on the ratio in the balanced equation, the limiting reactant is N2 since it produces fewer moles of NH3 compared to H2. From the balanced equation, 1 mole of N2 produces 2 moles of NH3.

Therefore, 0.89375 moles of N2 will produce (0.89375 moles * 2 moles of NH3/1 mole of N2) = 1.7875 moles of NH3.

Now, let's compare the volume of the initial mixture with the volume of the NH3 produced. Since we know that 1 mole of ideal gas occupies 22.4L at STP, we can calculate the volume of NH3 produced.

Volume of NH3 = 1.7875 moles * 22.4L/mol = 39.9875L

Therefore, the decrease in volume is:
Initial volume - Final volume = (20L + 100L) - 39.9875L = 80.0125L

So, the decrease in volume of the mixture is 80.0125L.

To determine the decrease in the volume of the mixture during the reaction, we need to calculate the difference between the initial volume and the final volume.

Given:
Initial volume of N2 (V(N2)) = 20 L
Initial volume of H2 (V(H2)) = 100 L

According to the balanced chemical equation:
N2 + 3H2 → 2NH3

From the balanced equation, we can deduce that:
1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

Therefore, we need to determine the number of moles for N2 and H2 in the given volumes and use the stoichiometry of the reaction to find the volume of NH3 produced.

To calculate the number of moles, we can use the ideal gas law equation:
n = PV / RT

Where:
n = number of moles
P = pressure (assumed constant)
V = volume
R = ideal gas constant
T = temperature (assumed constant)

To simplify the calculation, we assume constant pressure and temperature for all gases, so the equations can be simplified as:
n(N2) = V(N2) / 22.4 L/mol
n(H2) = V(H2) / 22.4 L/mol

Now, let's calculate the number of moles of N2 and H2:
n(N2) = 20 L / 22.4 L/mol ≈ 0.893 mol
n(H2) = 100 L / 22.4 L/mol ≈ 4.464 mol

According to the stoichiometry of the reaction, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the limiting reactant is N2, as it is present in the smallest amount.

Since each mole of N2 produces 2 moles of NH3, the number of moles of NH3 produced can be calculated as:
n(NH3) = [2 moles of NH3 / 1 mole of N2] * n(N2)
n(NH3) = 2 * 0.893 mol ≈ 1.786 mol

Now, we calculate the volume of NH3 produced using the same equation:
V(NH3) = n(NH3) * 22.4 L/mol
V(NH3) = 1.786 mol * 22.4 L/mol ≈ 39.97 L

To determine the decrease in volume, we subtract the initial volume of the mixture from the final volume of NH3 formed:
Decrease in volume = V(N2) + V(H2) - V(NH3)
Decrease in volume = 20 L + 100 L - 39.97 L ≈ 80.03 L

Therefore, the decrease in volume of the mixture is approximately 80.03 L.