Answer:
The answer to your question is a) 51.07 g of NiBr₂ b) Nickel, 54 g
Explanation:
Data
mass of NiBr₂ = ?
mass if Ni = 67.8 g
mass of Br = 37.3 g
Balanced chemical reaction
Ni + Br₂ ⇒ NiBr₂
Process
1.- Find the atomic mass of the reactants and the molar mass of the product
Ni = 59 g
Br = 79.9 x 2 = 159.8 g
NiBr₂ = 59 + 159.8 = 218.8 g
2.- Find the limiting reactant
theoretical yield Ni/Br₂ = 59/159.8 = 0.369
experimental yield Ni/Br₂ = 67.8/37.3 = 1.81
The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.
3.- Calculate the mass of NiBr₂
159.8 g of Br₂ --------------- 218.8 g of NiBr₂
37.3 g of Br₂ -------------- x
x = (37.3 x 218.8) / 159.8
x = 8161.24/159.8
x = 51.07 g of NiBr₂
4.- Find the excess reactant
The excess reactant is Nickel
59 g of Ni ---------------- 159.8 g of Br₂
x ---------------- 37.3 g of Br₂
x = (37.3 x 59)/159.8
x = 2200.7/159.8
x = 13.77 g of Ni
Excess Ni = 67.8 - 13.77
= 54 g
