The volume of 0.162 mna3po4 solution is necessary to completely react with 91.4 ml of 0.104 m cacl2 is 0.137 L
To determine the volume of 0.162 M Na3PO4 solution that is necessary to completely react with 91.4 mL of 0.104 M CaCl2, you can use the concept of stoichiometry..
The chemical reaction between Na3PO4 and CaCl2 can be represented by the following balanced chemical equation:
3 Na3PO4 + 2 CaCl2 -> Ca3(PO4)2 + 6 NaCl
First, you need to convert the volumes of the solutions to moles using the molarities and the volume/molarity formula:
moles Na3PO4 = (0.162 M) * (91.4 mL / 1000 mL/L) = 0.0150 moles
moles CaCl2 = (0.104 M) * (91.4 mL / 1000 mL/L) = 0.0095 moles
Next, use the molar ratios in the balanced chemical equation to determine the volume of Na3PO4 solution that is necessary to react with 0.0095 moles of CaCl2:
(Volume Na3PO4 / 3) = (Volume CaCl2 / 2)
Solving for the volume of Na3PO4, you get:
Volume Na3PO4 = (3 / 2) * (Volume CaCl2)
Plugging in the values, you get:
Volume Na3PO4 = (3 / 2) * (91.4 mL / 1000 mL/L)
Solving, you get:
Volume Na3PO4 = 0.137 L
Therefore, the volume of 0.162 M Na3PO4 solution that is necessary to completely react with 91.4 mL of 0.104 M CaCl2 is approximately 0.137 L.
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