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?

See explanation below for answers

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

a.

b.

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:

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:

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

b. Now, the excess moles of oxygen are:

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