2NAN3(s) 2Na(s) + 3N2(g)?

Question

I have to use that problem to answer this homework question and I really don't understand the questions. They are:
A. Write the six mole ratio that can be obtained from this equation
B. How many moles N2 can be formed from 2.7 mol NaN3
C. How many grams Na can be made from 31.1 g NaN3?
D. How many molecules N2 are made when 3.24 X 10 to the power of 23 atoms are formed?

My teacher said whoeve gets this question right will win a free homework pass for 3 weeks. I really need this because I am always made fun of and really would impress many if I got the correct answers. Help me out people if you can. Take care!

Answer 1

A. I don't know what you mean by "six mole ratio"
B. For each mole of NaN3 available 3/2 moles of N2 are formed.
=> 3/2 * 2.7 moles = 4.05 moles
C. The molecuar mass of NaN3 is 22.99 + 3*14.01 = 65.02. The part of Na in this is 22.99/65.02 => 31.1 g * 22.99/65.02 = 11.0 g
D. For each molecule N2 their are two atoms N. => 3.24 x 10^23/2 =1.62 x 10^23 molecules.