Consider the following chemical equation 2Na3PO4(aq) + 3CuCl2(aq) -> Cu3(PO4)2(s)+6NaCl (aq) a. What volume in mL...

Question

Question

Consider the following chemical equation 2Na3PO4(aq) + 3CuCl2(aq) -> Cu3(PO4)2(s)+6NaCl (aq)

a. What volume in mL of .1950M Na3PO4 is necessary to completly react with 85,6mL of .205 M CuCl2?

b. If 2.100 g are actually produced in an experiment what is the % reaction yield?

Answer 1

Answer #1

a) .205 M = x mol of CuCl₂ / .0856L

.01755 mol of CuCl₂ = 3

x mol of Na₃PO₄= 2

.0117 mol of Na₃PO₄ / L = .195M

V = 60 mL of Na₃PO₄

b) If we take as a theorical yield 1 mol, the yield is = 2.1 g / 380.58 g = .55%

If we take the previous experiment: 2.1g / 2.2264 g = 94.32%

Answer 2

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