Respuesta :
The volume of 0.175 M Na3PO4 solution that is necessary to completely react withn 95.4 ml of 0.102 M cuCl2 is
calculation
Step 1: write the equation for reaction
3CuCl2 (aq) + 2Na3Po4 → Cu3(PO4)2 + 6 NaCl
Step 2 : find the moles of CuCl2
moles = molarity x volume in liters
volume in liters = 95.4 /1000 =0.0954 L
moles is therefore = 0.0954 L x 0.102 M= 0.00973 moles
Step : use the mole ratio to calculate the moles of Na3Po4
The mole ratio of CuCl2 : Na3PO4 is 3:2 therefore = 0.00973 x2/3= 0.00649 moles
Step 4 : find the volume of Na3PO4
volume = moles/molarity
= 0.00649 mole / 0.175=0.037 M
0.073 L of 0.175 M [tex]\rm \bold{Na_3PO_4}[/tex] solution is necessary to completely react with 95.4 ml of 0.102 m [tex]\rm \bold{CuCl_2}[/tex].
The given reaction,
[tex]\rm \bold{3CuCl_2 (aq) + 2Na_3PO_4 \rightarrow Cu_3(PO4)_2 + 6 NaCl}[/tex]
3 molecules of copper chloride react with 2 sodium phosphate to form 1moles of copper phosphate.
Moles of [tex]\rm \bold{CuCl_2}[/tex],
[tex]\rm \bold{ n = M \times V}[/tex]
where,
M = molarity = 0.102
V = volume =95.4 /1000 = 0.0954 L
put the value we get
n = 0.00973 moles
Moles of [tex]\rm \bold{Na_3PO_4}[/tex]
n = 0.00973 x2/3
n = 0.00649 moles
Solve the molarity equation for volume, we get
V = 0.073 L
Hence, we can conclude that 0.073 L of 0.175 M [tex]\rm \bold{Na_3PO_4}[/tex] solution is necessary to completely react with 95.4 ml of 0.102 m [tex]\rm \bold{CuCl_2}[/tex].
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