Question

How much energy in kJ is released when 124.49 g of nickel react with an excess of phosphorus?

5 Ni + 2 P -> Ni₅P₂  

△ H= 436 kJ/mol

Answer 1

n = m / M

M (Ni) = 58.69 g/mol

n (Ni) = 124.49 / 58.69 = 2.12 mol

Q = 436 x 2.12 = 925 kJ