Given the following reaction: 3CuCl2(aq) 2Na3PO4(aq) --> Cu3(PO4)2(s) 6NaCl(aq) MM (g/mol) 134.45 163.94 380.58 58.44 If 285 mL of 6.3 M CuCl2 is added to excess Na3PO4 solution, how much precipitate( in grams) is produced Note: Write answer to two decimal places.

227.78g of the precipitate are produced

Explanation:

Based on the reaction, 3 moles of CuCl2 produce 1 mole of Cu3(PO4)2 (The precipitate).

To solve this question we need to find the moles of CuCl2 added. With these moles and the reactio we can find the moles of Cu3(PO4)2 and its mass as follows:

Moles CuCl2:

285mL = 0.285L * (6.3mol / L) = 1.7955 moles CuCl2

Moles Cu3(PO4)2:

1.7955 moles CuCl2 * (1mol Cu3(PO4)2 / 3mol CuCl2) = 0.5985 moles Cu3(PO4)2

Mass Cu3(PO4)2 -380.58g/mol-

0.5985 moles Cu3(PO4)2 * (380.58g/mol) =

227.78g of the precipitate are produced