How much of the excess reactant remains unchanged? how much N2 is formed? (in mol)
how much H2O is firmed? (in mol)

1. The limiting reactant is H2O2.

2. The amount of excess reactant, N2H4 that remain unchanged is 11.15 moles.

3. 1.65 moles of N2.

4. 6.6 moles of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

N2H4 + 2H2O2 —> N2 + 4H2O

From the balanced equation above,

1 mole of N2H4 reacted with 2 moles of H2O2 to produce 1 mole of N2 and 4 moles of H2O.

1. Determination of the limiting reactant.

From the balanced equation above,

1 mole of N2H4 reacted with 2 moles of H2O2.

Therefore, 12.8 moles of N2H4 will react with = 12.8 x 2 = 25.6 moles of H2O2.

From the calculations made above, we can see that it will take a higher amount i.e 25.6 moles than what was given i.e 3.3 moles of H2O2 to react completely with 12.8 moles of N2H4.

Therefore, H2O2 is the limiting reactant and N2H4 is the excess reactant.

2. Determination of the excess reactant that remain unchanged.

The excess reactant is N2H4.

First, we shall determine the amount of the excess reactant that reacted. This is illustrated below:

From the balanced equation above,

1 mole of N2H4 reacted with 2 moles of H2O2.

Therefore, Xmol of N2H4 will react with 3.3 moles of H2O2 i.e

Xmol of N2H4 = (1 x 3.3)/2

Xmol of N2H4 = 1.65 moles

Therefore, 1.65 moles of N2H4 reacted.

Now, we shall determine the excess reactant that remain unchanged. This can be obtained as follow:

Amount of N2H4 given = 12.8 moles

Amount of N2H4 that reacted = 1.65 moles.

Amount of N2H4 that remain unchanged =?

Amount of N2H4 that remain unchanged = (Amount of N2H4 given) – (Amount of N2H4 that reacted = 1.65 moles)

Amount of N2H4 that remain unchanged = 12.8 – 1.65

Amount of N2H4 that remain unchanged = 11.15 moles.

3. Determination of the amount of N2 produced.

In this case, the limiting reactant will be used because it will give the maximum yield of N2 as all of it is used up in the reaction.

The limiting reactant is H2O2 and the amount of N2 produced can be obtained as follow:

From the balanced equation above,

2 moles of H2O2 reacted to produce 1 mole of N2.

Therefore, 3.3 moles of H2O2 will react to produce = (3.3 x 1)/2 = 1.65 moles of N2.

Therefore, 1.65 moles of N2 were obtained from the reaction.

4. Determination of the amount of H2O produced.

In this case, the limiting reactant will be used because it will give the maximum yield of H2O as all of it is used up in the reaction.

The limiting reactant is H2O2 and the amount of H2O produced can be obtained as follow:

From the balanced equation above,

2 moles of H2O2 reacted to produce 4 moles of H2O.

Therefore, 3.3 moles of H2O2 will react to produce = (3.3 x 4)/2 = 6.6 moles of H2O.

Therefore, 6.6 moles of H2O were produced from the reaction.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa