What minimum mass of copper (II)nitrate must be added to 30.0 mL of a 0.0387 M phosphate solution in order to completely precipitate all of the phosphate as solid copper (II)phosphate? 2PO₄^3-(aq)+ 3Cu(NO₃)₂(aq)$\to$ Cu₃(PO₄)₂(s)+ 6NO₃^-(aq)

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Quizplus · Accepted Answer

First, calculate moles of phosphate: 0.0387 M * 0.030 L = 0.001161 mol. From the balanced equation, 2 moles of phosphate react with 3 moles of copper(II) nitrate, so 0.001161 mol of phosphate requires 0.001742 mol of copper(II) nitrate. The molar mass of Cu(NO3)2 is approximately 187.56 g/mol, so the mass needed is 0.001742 mol * 187.56 g/mol = 0.327 g.

What Minimum Mass of Copper (II)nitrate Must Be Added to 30.0