Final answer:
To find the mass of the precipitate that will form when 18.0 mL of a 0.334 M sodium phosphate solution reacts with 19.6 mL of a 0.155 M lead(II) nitrate solution, calculate the number of moles of each reactant and use the mole ratio to find the number of moles of the precipitate that will form. Finally, convert the moles of the precipitate to grams using its molar mass.
Step-by-step explanation:
In this reaction, lead (II) nitrate and sodium phosphate react to form lead (II) phosphate and sodium nitrate. The balanced equation for the reaction is:
2Na3PO4(aq) + 3Pb(NO3)2(aq) → Pb3(PO4)2(s) + 6NaNO3(aq)
By examining the stoichiometry of the reaction, we can determine that for every 3 moles of Pb(NO3)2, 1 mole of Pb3(PO4)2 will form. This means that the mole ratio of Pb(NO3)2 to Pb3(PO4)2 is 1:1/3.
To find the mass of the precipitate that will form, we can use the given volumes and concentrations of the solutions to calculate the number of moles of Pb(NO3)2 and Na3PO4. We can then use the mole ratio to find the number of moles of Pb3(PO4)2 that will form. Finally, we can convert the moles of Pb3(PO4)2 to grams using its molar mass.