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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?

Respuesta :

danidg02

Answer:

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

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