<span>Mr = 13 g / mol
mass = 5 g
Mol = 5/13 mol :)</span>
I am almost positive that the answer is D
Answer: 4.
and 
Explanation:
a) The given reaction is 
As the mass on both reactant and product side must be equal:


As the atomic number on both reactant and product side must be equal:



b) 
Total mass on reactant side = total mass on product side
15 =15 + x
x = 0
Total atomic number on reactant side = total atomic number on product side
8 = 7 + y
y = 1

ADD THEM all, and then divide by four. Thats what I would do!
Single displacement and combustion reactions are ALWAYS redox.