What volume of 0.149 M K3PO4 is required to

react with 41 mL of 0.793 M MgCl2 according
to the equation
2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl
Answer in units of m

Surely you want the answer in units of M and not m.

Here is a worked example of a stoichiometry problem. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To find the volume of 0.149 M K3PO4 required, we first need to determine the number of moles of MgCl2 reacted using the given concentration and volume.

Step 1: Convert the given volume of MgCl2 to moles.
Using the formula:
moles = concentration × volume
moles of MgCl2 = 0.793 M × 41 mL * (1 L / 1000 mL) = 0.032473 moles

Step 2: Determine the stoichiometric ratio between MgCl2 and K3PO4.
According to the balanced equation:
2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl
The ratio between the coefficients of MgCl2 and K3PO4 is 3:2.

Step 3: Calculate the moles of K3PO4 required.
Using the stoichiometric ratio, we can determine the moles of K3PO4 reacted:
moles of K3PO4 = (moles of MgCl2) × (2 moles of K3PO4 / 3 moles of MgCl2)
moles of K3PO4 = 0.032473 moles × (2/3) = 0.021649 moles

Step 4: Calculate the volume of 0.149 M K3PO4 required.
Now we can use the moles of K3PO4 calculated in the previous step to find the volume using the formula:
volume = moles / concentration
volume of K3PO4 = 0.021649 moles / 0.149 M = 0.14555 L

Answer:
The volume of 0.149 M K3PO4 required to react with 41 mL of 0.793 M MgCl2 is approximately 0.14555 liters (L).