What is the maximum mass of cobalt(II) phosphate that will precipitate from a so
ID: 1053916 • Letter: W
Question
What is the maximum mass of cobalt(II) phosphate that will precipitate from a solution prepared by adding an excess of a K_3PO_4 solution to 100.0 mL of 5.0 M CoCl_2? Bonus: You are sitting in a quiet room working on your chemistry take-home exam. All of a sudden you feel panic come over you and thus you decide to get up and take a break to get some fresh air. When you try to leave, you learn that only door out of the room is locked. Assuming there is no air flow into the room (it is air tight with no ventilation system) How long can you survive in this room (5 meters long. 5 meters wide. 5 meters high) with your chemistry exam? Keep in mind the following facts (you may no. need to use all of them depending on how you approach this problem), and feel, free to research any information that, you think is necessary to answer this question, but remember to cite any outside information. Human tidal volume (the volume of air that moves In and out of the lungs during normal quiet respiration) is about 500 mL for an adult. Resting respiratory rate for adults is 12-18 breaths per minute (bpm). Under stressfulExplanation / Answer
Given that 100 mL of 5M CoCl2.
Molar mass of CoCl2 = 129.8 g/mol
Molar mass of Co3(PO4)2 = 366.7 g/mol
We know that
Molarity = (mass/ molar mass) x ( 1/ volume of solution in Litres)
Hence,
Mass of CoCl2 = Molarity x molar mass x volume of solution in Litres
= 5 M x 129.8 g/mol x 0.1 L
= 64.9 g
3CoCl2 + 2 K3PO4 ------------------> Co3(PO4)2 + 6KCl
3 mol 1 mol
3 x 129.8 g = 389.4 g 366.7 g
64.9 g ?
? = ( 64.9 g/ 389.4 g) x 366.7 g of Co3(PO4)2
= 61.12 g of Co3(PO4)2
Therefore,
Maximum mass of Co3(PO4)2 that can be precipitated = 61.12 g
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