What volume of 0.195 M N a 3 P O 4 solution is necessary to completely react wit
ID: 889078 • Letter: W
Question
What volume of 0.195 M Na3PO4 solution is necessary to completely react with 87.0 mL of 0.117 M CuCl2?Exercise 4.65 Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq) What volume of 0.195 M Na3PO4 solution is necessary to completely react with 87.0 mL of 0.117 M CuCl2?
Exercise 4.65 Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq)
Exercise 4.65 Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq) Exercise 4.65 Exercise 4.65 Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq) Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq) Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq)
Explanation / Answer
2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq)
No of moles of CuCl2 = 0.117*87/1000 = 0.0102 mole
from equation 2 mole Na3PO4 = 3 mole CuCl2
so that , No of moles of Na3PO4 required = 0.0102*2/3 = 0.0068 mole
therefore,
volume of 0.195 M Na3PO4 solution required = n/M = 0.0068 / 0.195
= 0.035 L
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