Limiting Reactants Question?

Question

1. Some rocket engines use a mixture of hydrazine, N2 H4, and hydrogen peroxide, H2 O2, as the propellant. The reaction is given by the following equation.

N2H4 (l) + 2H202 (l) ---> N2 (g) + 4H2O (g)

Side question: Why does the need of (l) liquid or (g) or aq(which is aqueous, dissolving in water) need to be specified. I've been doing problems with these signs since I started balancing redox reactions several chapters in general chemistry.. it hasn't seemed to affect the final answer.

a. Which is the limiting reactant in this reaction when 0.750 mol N2H4 is mixed with 0.500 mol H2O2?

b. How much of the excess reactant, in moles, remains unchanged?

c. How much of each product, in moles, is formed/

My work:
a. answer: 0.750 mol N2H4 produces 3 mol H20 ( I just looked at the ratios of H20 produced from each amount. )

b. I don't know how to find that. (????)

c. I put 0.25 mol N2, and 1 mol H20 (from plug, but I don't see why you can't plug in 0.75 mol N2H4 to find this answer)

Answer 1

a. The limiting reagent is the 0.500 mole of H2O2, not the 0.750 mole of N2H4. 0.750 mole of N2H4 requires 1.500 moles of H2O2 . 0.500 mole of H2O2 requires only 0.250 mole of N2H4

b. How much of the excess reactant, in moles, remains unchanged?
0.500 mole of N2H4 remains unchanged.

c. How much of each product, in moles, is formed?
0.250 mole of N2 (g) + 1.000 mole of H2O (g)

You can't use the 0.750 because there is insufficient H2O2 to oxidize it.

Specifying the condition of the reagents and the products gives clues on equilibrium and reversibility. In this reaction since the products are gases there can be no equilibrium unless the reaction is contained at very high pressure.

Answer 2

a. The ratio between N2H4 and H2O2 is 1 : 2 so we would need 0.750 x 2 = 1.5 moles of H2O2. Since we have only 0.500 mol of H2O2 it is the limiting rewactant and N2H4 is in excess.

b. The ratio is 1 : 2
moles N2H4 needed to react with 0.500 mol of N2O2 = 0.500/2 = 0.250
Moles N2H4 in excess = 0.750 - 0.250 = 0.500

c.To answer at this question we have to work on the limiting reactant.
the ratio between H2O2 ( limiting reactant ) and N2 is 2 : 1
moles N2 produced = 0.5 / 2 = 0.250
the ratio between H2O2 and H2O is 2 : 4
moles H2O produced = 0.500 x 4 / 2 = 1.0

Answer 3

while balancing the redox equations states need not be specified .But in the lab N2H4 may be obtained in three states .which one to use? so specification is done.
See 2 moles H2O2 reacts with 1mole N2H4. therefore 1mole reacts with.5 moles N2H4. but H2O2 given is .5moles which reacts with .25 moles N2H4.therefore N2H4 remains in excess and .5 of it remains.H2O2 is the limiting reagent.
1 mole N2H4 reacts to form 1 mole N2 therefore .25 of it react to form N2=.25moles.1mole N2H4 produces 4 moles h20 therefore .25 moles produce .25*4=1 mole H2O.

Answer 4

Since the equations tell you the number of MOLES of something that are reacting and being produced you have to convert your grams into moles so you can compare. The first equation tells you that 1 mol of KOH will react with 1 mol of HNO3. So do you have a 1 : 1 mol ratio with the grams given? Let's see: 16 g of KOH = 16 g / 56.11 g/mol = 0.285 mol 12 g of HNO3 is 12 g / 63.02 g/mol = 0.190 Well it appears you DON'T have equal numbers of moles. there are more moles of KOH than HNO3. So the reaction will proceed but only until 0.190 mol of HNO3 is used up and 0.190 mol of KOH is used, then the reaction will stop. YOu will end up with (0.285 mol - 0.190 mol = 0.095 mol) of KOH left over not reacted. The HNO3 is the LIMITING REAGENT then because it limits how far the reaction can go before it stops. The second equation is a little more complicated but not much. mol of NaOH = 10 g / 40 g/mol = 0.250 mol mol of H2SO4 is 10 g / 98.08 g/mol = 0.102 mol Now from the H2SO4's point of view, if it wants to be all used up it will need to react with TWICE as many moles of NaOH. Notice the equation says that? ratio is 2 : 1 So what is 0.102 x 2? well, it is 0.204 mol. And we have enough NaOH. Much more than enough. So once again the reaction will proceed until 0.102 mol of H2SO4 is used up then stop. So the H2SO4 is the limiting reagent, and when the reaction stops there will be a littl unused NaOH. Get It?

Answer 5

1. In the early days, most chemistry was aqueous. Most redox reactions are in water, since you are using H+ or H2O to supply the surplus in electrons or oxygen to the reactions. But in this reaction, the liquids ARE hydrazine and H2O2. As for the reaction, you missed the trees and got lost in the forest. If you have 0.750 moles of hydrazine, 0.375 moles of peroxide are sufficient to complete the reaction. Since you have more, it is the hydrazine that is limiting.
From the discussion above, 0.125 moles of peroxide are unreacted. From the reaction coefficients, 0.375 moles of peroxide produce 0.750 moles of water, and the 0.750 moles of hydrazine that react produce 0.750 moles of N2.

Answer 6

Hydrogen Peroxide Propellant