Answer A and B **Problem Overview:**A solution contains an unknown amount of dissolved calcium. The addition of 0.352 mol of K₃PO₄ causes complete precipitation of all the calcium.**Part A:**- Question: How many moles of calcium were dissolved in the solution?- Answer box with variable "ν =" followed by an input field labeled "moles."**Part B:**- Question: What mass of calcium was dissolved in the solution?- Answer box with variable "m =" followed by an input field labeled "g."Each part provides functions for mathematical input, undo/redo options, and a seating chart icon for additional help. There are "Submit" and "Request Answer" buttons available for each part.

For solving this problem first we have to write the balanced chemical equation for the reaction 2K3PO4(s) + 3Ca2+(aq) →Ca3(PO4)2 + 6K+ Here we can say that 0.352 mole of K3PO4 creates 0.352 mol (PO4 )2- From the above equation, we can say that for every 2 moles of K3PO4 , 3 moles of calcium consumed So the moles of calcium dissolved in the solution can be calculated as 0.352 mol K3PO4 X 3 mol Ca2 mol K3PO4 = 0.528 mol Ca so moles of calcium dissolved = 0.528 mol Mass of calcium dissolved in the solution can be calculated by multiply moles of calcium dissolved with molar mass of calcium molar mass of calcium =40.078 g/mol so mass of calcium dissolved = 0.528 mol X 40.078 g/mol =21.161 g Ca