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
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
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