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

1 answer

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.
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