posted by Vinny on .
What volume of .300 M CaCl2 is needed to react completely with 40.00 Ml of .200 M H2PO4. according to the following equation
3 CaCl2 (aq) + 2 H3PO4 (aq)-> Ca3(po44)2 + 6 HCL (aq)
you need 3/2 moles of CaCl2 for each mole of H3PO4.
moles of phosphoric acid: .04*.2
moles of calcium chloride: 3/2 * .008
moles of calcium chloride= volume*.3
volume= (3/2 .008)/.3 liters
If the 3 moles of Calcium Nitride was then dissolved in 2 L of liquid ammonium what would the molarity of the Calcium Nitride /ammonium solution be?