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In a chemical reaction, the limiting reactant is the substance that determines the amount of product that can be formed. It is entirely consumed when the reaction goes to completion. Once the limiting reactant is used up, the reaction stops, and no more products can form. This concept is crucial because it allows chemists to predict the maximum yield of a reaction.
To identify the limiting reactant, compare the mole ratios of the reactants based on the balanced equation. For the reaction between sodium phosphate (Na鈧働O鈧�) and lead(II) acetate [Pb(CH鈧僀OO)鈧俔, we need to use their stoichiometric coefficients.

• The balanced equation is: 2Na鈧働O鈧� + 3Pb(CH鈧僀OO)鈧� 鈫� Pb鈧�(PO鈧�)鈧� + 6Na(CH鈧僀OO).
• Ensure that the correct mole ratio is maintained as per the equation. Here, 2 moles of Na鈧働O鈧� react with 3 moles of Pb(CH鈧僀OO)鈧�.
• Calculate the mole-to-coefficient ratio for both reactants and identify which one has a smaller value.
• In this case, Pb(CH鈧僀OO)鈧� with a ratio of 0.0133 is the limiting reactant.

Recognizing the limiting reactant helps in efficiently planning chemical reactions to avoid waste and maximize yield.

The theoretical yield is the maximum amount of product that can be generated from a chemical reaction, assuming complete conversion of the limiting reactant into the desired product. Calculating the theoretical yield is vital for understanding the efficiency of a reaction.
To find the theoretical yield:

• Use the balanced chemical equation to determine the stoichiometry of the reaction.
• Multiply the moles of the limiting reactant by the stoichiometry factor from the balanced equation to find the moles of the product.
• Then convert these moles of product into grams using the molar mass of the product.

For the reaction at hand, 3 moles of Pb(CH鈧僀OO)鈧� yield 1 mole of Pb鈧�(PO鈧�)鈧�. With Pb(CH鈧僀OO)鈧� as the limiting reactant:

• 0.0400 moles of Pb(CH鈧僀OO)鈧� provide \(\frac{1}{3}\) times that in moles of Pb鈧�(PO鈧�)鈧�, which equates to 0.0133 moles.
• Using Pb鈧�(PO鈧�)鈧�'s molar mass of 811.1 g/mol, the mass of the precipitate is calculated to be 10.8 g.

This shows that if the reaction goes perfectly, you can expect 10.8 grams of lead phosphate (Pb鈧�(PO鈧�)鈧�) as your theoretical yield.

Molar concentration, often expressed in moles per liter (M), indicates the number of moles of a solute present in one liter of solution. This is a fundamental concept in chemistry, as it helps to understand how much of a substance is dissolved in a solution.
To calculate molar concentration after a reaction, especially to find the concentration of any excess reactant, consider:

• The initial amount of the reactant present before the reaction.
• The volume of the solution after the reaction takes place.

In the case of the initial problem, sodium phosphate (Na鈧働O鈧�) is the excess reactant. After the reactants have reacted:

• The excess moles of Na鈧働O鈧� remaining are 0.0533 moles.
• Considering a final solution volume of 150.0 mL (0.150 L), use the formula: molar concentration = moles / volume.
• This results in a concentration of 0.355 M for PO鈧劼斥伝 ions.

This calculation of molar concentration is essential in ensuring proper reactant usage and for the potential application of the solution in further processes or reactions.