What volume of 0.596M K3PO4 is required to
react with 29 mL of 0.864 M MgCl2 according to the equation
2K3PO4 + 3MgCl2 -> Mg3(PO4)2 + 6KCl
Answer in units of mL
AP Chemistry - DrBob222, Sunday, October 21, 2012 at 8:26pm
mols MgCl2 = M x L = ?
Use the coefficient in the balanced equation to convert mols MgCl2 to mols K3PO4.
Then M K3PO4 = mols K3PO4/L K3PO4. You know M and mols, solve for L and colnvert to mL.