Question

An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. balanced chemical equation: Pb(NO3)2(aq)+2KCl(aq)⟶PbCl2(s)+2KNO3(aq)Pb(NO3)2(aq)+2KCl(aq)⟶PbCl2(s)+2KNO3(aq) What is the limiting reactant? potassium chloride lead(II) nitrate The percent yield for the reaction is 87.5%87.5% . How many grams of the precipitate are formed? precipitate formed: gg How many grams of the excess reactant remain?

Answer 1

Answer:

• The limiting reactant is lead(II) nitrate.

• 7.20 g of precipitate are formed.

• 1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

• Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we convert the masses of each reactant to moles, using their molar mass:

• 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂

• 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we use the moles of the limiting reactant:

• 0.0296 mol Pb(NO₃)₂ $\frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}}$ * $\frac{278.1g}{1molPbCl_{2}}$ * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

• 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we convert that amount to moles of KCl and finally into grams of KCl:

• 0.0259 mol Pb(NO₃)₂ $\frac{2molKCl}{1molPb(NO_{3})_2}$ * $\frac{74.55g}{1molKCl}$ = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

• 5.76 - 3.86 = 1.9 g KCl