Consider the following precipitation reaction: 2Na3PO4 (aq) + 3CuCl2 (aq) → Cu3 (PO4) 2 (s) + 6NaCl (aq)

What volume of 0.184 M Na3PO4 solution is necessary to completely react with 94.6 mL of 0.108 M CuCl2?

V = 37.0 mL

Explanation:

First find the moles of the known substance (CuCl2)

n = cv

where

n is moles

c is concentration

v is volume (in litres)

n = 0.108 * 0.0946

n=0.0102168

Using the mole ratio in the balanced reaction, we can find the moles of Na3PO4

n (Na3PO4) = n (CuCl2) * 2/3

=0.0102168 * 2/3

=0.0068112

Now we have all the necessary values to calculate the volume

v=n/c

v = 0.0068112/0.184

v = 0.0370173913 L

v = 37.0 mL