3Mg + N₂= Mg₃N₂
n(Mg)=12,2g÷<span>24,4g/mol=0,5mol-limiting reagent
</span>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%.
The answer is 232 plus 450
_Mg + _HCL = _MgCl2 + H2
Separate the terms on each side:
_Mg + _HCl = _MgCl2 + H2
Mg- 1 Mg-1
H-1 H-2
Cl-1 Cl-2
Mg is balanced on both sides so move on to the next (put a 1 in the space).
1Mg
There are two H's and two Cl's on the results side, so to balance the equation put a 2 as a coefficient for HCl and it'll all balance out.
2HCl
Balamced equation will be:
1Mg + 2HCL = 1MgCl2 + H2
The second thesis statement is perfect. It supports the claim and presents main idea.
