Question

Hydrazine (N2H4) is used as rocket fuel. It reacts with oxygen to form nitrogen and water.

N2H4(l) + O2(g)  N2(g) + 2H2O(g)


a. How many liters of N2 (at STP) form when 2.0 kg N2H4 reacts with 2.1 kg O2?

b. How many grams of the excess reagent remain after the reaction?

Answer 1

a.
`V_(N_2)=1399.1L`

b.
`m_(O_2)^(excess)=0.1kg`

Step-by-step explanation:

Hello,

a. In this case, we first identify the limiting reactant by computing the available moles of hydrazine and the moles of hydrizine that 2.1 kg of oxygen would consume:

`n_(N_2H_4)^(available)=2000 gN_2H_4*(1molN_2H_4)/(32gN_2H_4) =62.5molN_2H_4\\n_(N_2H_4)^(reacted\ with\ O_2)=2100gO_2*(1molO_2)/(32gO_2) *(1molN_2H_4)/(1molO_2) =65.625molN_2H_4`

Thus, since there are less available hydrazyne than it consumed, we state hydrazine is the limiting reactant, for that reason, the yielded moles of nitrogen are:

`n_(N_2)=62.5molN_2H_4*(1molN_2)/(1molN_2H_4) =62.5molN_2`

Next, by using the ideal gas equation at STP conditions (273 K and 1 atm) we compute the volume:

`V_(N_2)=(n_(N_2)RT)/(P)=(62.5mol*0.082(atm*L)/(mol*K)*273K)/(1 atm) \\ \\V_(N_2)=1399.1L`

b. Now, the excess moles of oxygen are:

`m_(O_2)^(excess)=65.625mol-62.5mol=3.125molO_2*(32gO_2)/(1molO_2) *(1kg)/(1000g)\\\\ m_(O_2)^(excess)=0.1kg`

Best regards.

Answer 2

See explanation below for answers

Step-by-step explanation:

This is a stochiometry reaction. LEt's write the overall reaction again:

N₂H₄ + O₂ ---------> N₂ + 2H₂O

This reaction is taking place at Standard temperature and pressure conditions (STP) which are P = 1 atm and T = 273 K. To know the volume of N₂ formed, we need to know first how many moles are formed, and this can be calculated with the reagents and the limiting reagent. Let's calculate the moles first of the reagents:

MM N₂H₄ = 32 g/mol; MM O₂ = 32 g/mol

mol N₂H₄ = 2000 / 32 = 62.5 moles

mol O₂ ? 2100 / 32 = 65.63 moles

Now that we have the moles, we need to apply the stochiometry and calculate the limiting reagent. According to the overall reaction we have a mole ratio of 1:1 between N₂H₄ and O₂, therefore:

1 mole N₂H₄ ---------> 1 mole O₂

62.5 moles ----------> X

X = 62.5 moles of O₂

But we have 65.63 moles, therefore, the limiting reactant is the N₂H₄.

We also have a 1:1 mole ratio with the N₂, so:

moles N₂H₄ = moles N₂ = 62.5 moles

Now that we have the moles, we can calculate the volume with the ideal gas equation:

PV = nRT

V = nRT / P

R: gas constant (0.082 L atm / K mol)

Replacing we have:

v = 62.5 * 0.082 * 273 / 1

V = 1399.13 L of N₂

Now, how many grams of the excess remains?, we know how many moles are reacting so, let's see how much is left:

moles remaining = 65.63 - 62.5 = 3.12 moles

then the mass of oxygen:

m = 3.12 * 32 = 100.16 g of O₂