mols MgCl2 = M x L = ?
Convert mols MgCl2 to mols K3PO4.
M K3PO(4 = mols K3PO4/L K3PO4. Solve for L and convert to mL.
What volume of 0.714 M K3PO4 is required to react with 27 mL of 0.97 M MgCl2 according to the equation
2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl Answer in units of mL
1 answer
1 answer