An aqueous solution containing 9.83 g of lead(II) nitrate is added to an aqueous solution containing 6.94 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) What is the limiting reactant? potassium chloride lead(II) nitrate PbCl2 (s) + 2KNO3(aq) The percent yield for the reaction is 87.2%. How many grams of the precipitate are formed?
The Correct Answer and Explanation is :
Balanced Chemical Equation:
The balanced chemical equation for the reaction between lead(II) nitrate and potassium chloride is:
[
\text{Pb(NO}_3\text{)}_2(aq) + 2\text{KCl}(aq) \rightarrow \text{PbCl}_2(s) + 2\text{KNO}_3(aq)
]
Determining the Limiting Reactant:
To determine the limiting reactant, we need to first calculate the number of moles of each reactant.
Moles of Lead(II) Nitrate (Pb(NO₃)₂):
The molar mass of lead(II) nitrate, Pb(NO₃)₂, is calculated as:
[
\text{Molar mass of Pb(NO}_3\text{)}_2 = 207.2 \, (\text{Pb}) + 2 \times (14.01 \, (\text{N}) + 3 \times 16.00 \, (\text{O})) = 331.2 \, \text{g/mol}
]
The moles of Pb(NO₃)₂ are:
[
\text{moles of Pb(NO}_3\text{)}_2 = \frac{9.83 \, \text{g}}{331.2 \, \text{g/mol}} = 0.0297 \, \text{mol}
]
Moles of Potassium Chloride (KCl):
The molar mass of potassium chloride (KCl) is:
[
\text{Molar mass of KCl} = 39.10 \, (\text{K}) + 35.45 \, (\text{Cl}) = 74.55 \, \text{g/mol}
]
The moles of KCl are:
[
\text{moles of KCl} = \frac{6.94 \, \text{g}}{74.55 \, \text{g/mol}} = 0.0931 \, \text{mol}
]
Stoichiometric Calculation:
From the balanced chemical equation, we know that 1 mole of Pb(NO₃)₂ reacts with 2 moles of KCl. Therefore, the required moles of KCl to completely react with 0.0297 mol of Pb(NO₃)₂ are:
[
\text{Required moles of KCl} = 2 \times 0.0297 = 0.0594 \, \text{mol}
]
Since we have 0.0931 mol of KCl, which is more than enough to react with the 0.0297 mol of Pb(NO₃)₂, lead(II) nitrate (Pb(NO₃)₂) is the limiting reactant.
Calculating the Theoretical Yield of PbCl₂:
From the balanced equation, 1 mole of Pb(NO₃)₂ produces 1 mole of PbCl₂. Therefore, 0.0297 moles of Pb(NO₃)₂ will produce 0.0297 moles of PbCl₂.
The molar mass of PbCl₂ is:
[
\text{Molar mass of PbCl}_2 = 207.2 \, (\text{Pb}) + 2 \times 35.45 \, (\text{Cl}) = 278.1 \, \text{g/mol}
]
The theoretical yield of PbCl₂ is:
[
\text{Theoretical yield of PbCl}_2 = 0.0297 \, \text{mol} \times 278.1 \, \text{g/mol} = 8.27 \, \text{g}
]
Calculating the Actual Yield:
The percent yield is 87.2%. Therefore, the actual yield of PbCl₂ is:
[
\text{Actual yield of PbCl}_2 = \frac{87.2}{100} \times 8.27 \, \text{g} = 7.22 \, \text{g}
]
Conclusion:
The limiting reactant is lead(II) nitrate (Pb(NO₃)₂), and the amount of precipitate formed, considering the 87.2% yield, is 7.22 grams.