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

Answer:

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

Answer 2

Answer:

a. $V_{N_2}=1399.1L$

b. $m_{O_2}^{excess}=0.1kg$

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:

$n_{N_2H_4}^{available}=2000 gN_2H_4*\frac{1molN_2H_4}{32gN_2H_4} =62.5molN_2H_4\\n_{N_2H_4}^{reacted\ with\ O_2}=2100gO_2*\frac{1molO_2}{32gO_2} *\frac{1molN_2H_4}{1molO_2} =65.625molN_2H_4$

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:

$n_{N_2}=62.5molN_2H_4*\frac{1molN_2}{1molN_2H_4} =62.5molN_2$

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

$V_{N_2}=\frac{n_{N_2}RT}{P}=\frac{62.5mol*0.082\frac{atm*L}{mol*K}*273K}{1 atm} \\ \\V_{N_2}=1399.1L$

b. Now, the excess moles of oxygen are:

$m_{O_2}^{excess}=65.625mol-62.5mol=3.125molO_2*\frac{32gO_2}{1molO_2} *\frac{1kg}{1000g}\\\\ m_{O_2}^{excess}=0.1kg$

Best regards.