Question

A rocket can be powered by the given reaction: N2O4(1) + 2N2H4(1) -> 3N2(g) + $H2O(g). An engineer designed the rocket to hold 1.00 kg of N2O4 and excess N2H4.

a)How much N2 would be produced with the 1.00 kg N2O4?

b)If 685g of N2 is obtained, what is the %yield of the process?

Answer 1

a)

Molar mass of N2O4,

MM = 2*MM(N) + 4*MM(O)

= 2*14.01 + 4*16.0

= 92.02 g/mol

mass(N2O4)= 1000 g

number of mol of N2O4,

n = mass of N2O4/molar mass of N2O4

=(1000.0 g)/(92.02 g/mol)

= 10.87 mol

Balanced chemical equation is:

N2O4 + 2 N2H4 ---> 3 N2 + 4 H2O

Molar mass of N2 = 28.02 g/mol

According to balanced equation

mol of N2 formed = (3/1)* moles of N2O4

= (3/1)*10.8672

= 32.6016 mol

mass of N2 = number of mol * molar mass

= 32.6*28.02

= 913.5 g

Answer: 913.5 g

b)

% yield = actual mass*100/theoretical mass

= 685*100/913.5

= 75.0 %

Answer: 75.0 %