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

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.

To determine the volume of 0.714 M K3PO4 required to react with 27 mL of 0.97 M MgCl2, we need to use the stoichiometry of the balanced equation.

The balanced equation is:
2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl

First, we need to determine the number of moles of MgCl2 in the given solution.
Number of moles of MgCl2 = concentration (M) x volume (L)
= 0.97 M x (27 mL/1000 mL) (converting mL to L)
= 0.02619 moles

According to the stoichiometry of the balanced equation, it states that 2 moles of K3PO4 react with 3 moles of MgCl2.

Using this information, we can set up a ratio:
(2 moles K3PO4 / 3 moles MgCl2) = (x moles K3PO4 / 0.02619 moles MgCl2)

Solving for x, the moles of K3PO4:
x = (2/3) * 0.02619 moles
x = 0.01746 moles

Now, we can find the volume of 0.714 M K3PO4 required using the previously calculated moles.
Volume (L) = moles / concentration (M)
Volume (L) = 0.01746 moles / 0.714 M
Volume (L) = 0.02443 L

Finally, we can convert the volume from liters to milliliters:
Volume (mL) = 0.02443 L * 1000
Volume (mL) = 24.43 mL

Therefore, the volume of 0.714 M K3PO4 required to react with 27 mL of 0.97 M MgCl2 is 24.43 mL.