How many grams of PbBr2 will precipitate when excess NH4Br solution is added to ...

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chionofalco @chionofalco

November 2022 1 142 Report

How many grams of PbBr2 will precipitate when excess NH4Br solution is added to 60.0 mL of 0.710 M Pb(NO3)2 solution?

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Tokamak

$\tt{\huge{\red{Solution:}}}$

Step 1: Write the balanced chemical equation.

Pb(NO₃)₂ + 2 NH₄Br → PbBr₂ + 2 NH₄NO₃

Step 2: Calculate the number of moles of Pb(NO₃)₂ using dimensional analysis.

Note that 1 M = 1 mol/L.

$\begin{aligned} \text{moles of} \: \text{Pb}(\text{NO}_3)_2 & = 60.0 \: \cancel{\text{mL soln}} \times \frac{1 \: \cancel{\text{L soln}}}{1000 \: \cancel{\text{mL soln}}} \times \frac{\text{0.710 mol} \: \text{Pb}(\text{NO}_3)_2}{1 \: \cancel{\text{L soln}}} \\ & = \text{0.0426 mol} \: \text{Pb}(\text{NO}_3)_2 \end{aligned}$

Final Step: Calculate the mass of PbBr₂ that will precipitate using dimensional analysis.

Based on the balanced chemical equation, 1 mole of Pb(NO₃)₂ is stoichiometrically equivalent to 1 mole of PbBr₂. Note that the molar mass of PbBr₂ is 367.0 g/mol.

$\begin{aligned} \text{mass of} \: \text{PbBr}_2 & = 0.0426 \: \cancel{\text{mol} \: \text{Pb}(\text{NO}_3)_2} \times \frac{1 \: \cancel{\text{mol} \: \text{PbBr}_2}}{1 \: \cancel{\text{mol} \: \text{Pb}(\text{NO}_3)_2}} \times \frac{\text{367.0 g} \: \text{PbBr}_2}{1 \: \cancel{\text{mol} \: \text{PbBr}_2}} \\ & = \boxed{\text{15.6 g} \: \text{PbBr}_2} \end{aligned}$

Hence, 15.6 g of PbBr₂ will precipitate.

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