Answer to Question #177651 in General Chemistry for kyle

Question

Consider the reaction between 55.0g of copper (II) nitrate tetrahydrate and 55.0 g ammonium phosphate. How many total oxygen atoms participated in the reaction?

Answer 1

3Cu(NO₃)₂*4H₂O + 2(NH₄)₃PO₄ = Cu₃(PO₄)₂ + NH₄NO₃ + 12H₂O

n((NH₄)₃PO₄) = m/Mr = (55 g)/(149 g/mol) = 0.3691 mol

n(Cu(NO₃)₂*4H₂O) = m/Mr = (55 g)/(260 g/mol) = 0.2115 mol - limiting reactant

Cu(NO₃)₂*4H₂O has 10 oxygen atoms per formula unit and (NH₄)₃PO₄ has 4 oxygen atoms per formula unit, so

n(O) = (0.2115 mol)*(10 + 2/3*4) = 2.679 mol

N(O) = n(O)*N_A = (2.679 mol)*(6.02*10²³ mol^-1) = 1.61*10²⁴ atoms