What minimum mass of cobalt (II)nitrate must be added to 60.0 mL of a 0.0999 M phosphate solution in order to completely precipitate all of the phosphate as solid cobalt (II)phosphate? 2PO₄^3-(aq)+ 3Co(NO₃)₂(aq)→ Co₃(PO₄)₂(s)+ 6NO₃^-(aq)

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The balanced chemical equation for the reaction is:$$ \text{2 PO}_4^{3-} (aq) + 3 \text{Co(NO}_3)_2 (aq) \rightarrow \text{Co}_3(PO_4)_2 (s) + 6 \text{NO}_3^- (aq) $$From the equation, 3 moles of cobalt(II) nitrate are needed to precipitate 2 moles of phosphate ions. First, calculate the moles of phosphate in the solution:$$ \text{Moles of phosphate} = 0.0999 \, \text{M} \times 0.0600 \, \text{L} = 0.005994 \, \text{mol} $$According to the stoichiometry of the reaction, the moles of cobalt(II) nitrate needed are:$$ \text{Moles of Co(NO}_3)_2 = \frac{3}{2} \times 0.005994 \, \text{mol} = 0.008991 \, \text{mol} $$The molar mass of cobalt(II) nitrate, Co(NO3)2, is approximately 182.94 g/mol (Co = 58.93, N = 14.01, O = 16.00). Therefore, the mass of cobalt(II) nitrate required is:$$ \text{Mass} = 0.008991 \, \text{mol} \times 182.94 \, \text{g/mol} = 1.645 \, \text{g} $$Thus, the minimum mass of cobalt(II) nitrate required is approximately 1.64 g.

What Minimum Mass of Cobalt (II)nitrate Must Be Added to 60.0