What volume of 0.300 M Na3PO4 is required to precipitate all the lead(II) ions from 130.0 mL of 0.400 M Pb(NO3)2?

Can you please show the steps I have no idea how to solve

I did that below for you and showed the steps. I shall be happy to answer any questions about any of the steps but show your work so I can analyze the problem.

I don't understand your steps if you can type them clearly that would be great, so far I have 0.400*0.13=5.2

What's next?

Sure! Let's break down the problem step by step.

Step 1: Write the balanced chemical equation for the reaction between Na3PO4 and Pb(NO3)2:
2 Na3PO4 + 3 Pb(NO3)2 -> 6 NaNO3 + Pb3(PO4)2

Step 2: Calculate the moles of Pb(NO3)2 in the given volume of solution:
Using the given concentration, we can calculate the moles of Pb(NO3)2:
Moles of Pb(NO3)2 = concentration x volume
Moles of Pb(NO3)2 = 0.400 M x 0.1300 L

Step 3: Calculate the molar ratio between Pb(NO3)2 and Na3PO4:
From the balanced equation, we can see that it takes a 3:2 ratio of Pb(NO3)2 to Na3PO4 for the complete reaction.

Step 4: Calculate the moles of Na3PO4 required:
Using the molar ratio, we can calculate the moles of Na3PO4 needed:
Moles of Na3PO4 = (Moles of Pb(NO3)2) x (2 moles Na3PO4 / 3 moles Pb(NO3)2)

Step 5: Calculate the volume of Na3PO4 needed:
Now we need to calculate the volume of 0.300 M Na3PO4 needed to provide the calculated moles:
Volume of Na3PO4 = (Moles of Na3PO4) / (concentration of Na3PO4)

Step 6: Substitute the values into the equation:
Substitute the calculated value of Moles of Pb(NO3)2 into the equation from step 4, then substitute the calculated value of Moles of Na3PO4 into the equation from step 5 to get the final answer.

Note: Make sure to unit match, for example, convert liters to milliliters if necessary.

I hope this explanation helps! Let me know if you have any further questions.