Question

A compound is formed when 12.2 g Mg combines completely with 5.16 g N. What is the percent composition of this compound? Please show all work and include the percentage of both magnesium and nitrogen.

Answer 1

3Mg + N₂= Mg₃N₂
n(Mg)=12,2g÷24,4g/mol=0,5mol-limiting reagent
n(N₂)=5,16g÷28g/mol=0,18mol
n(Mg₃N₂):n(Mg)=1:3, n(Mg₃N₂)=0,166mol, m(Mg₃N₂)=0,166·101,2=16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2=27,66%=0,2766
%(Mg) = 3·Ar(Mg)÷Mr(Mg₃N₂)= 3·24,4÷101,2=72,34% or 100% - 27,66%= 72,34%.