Question

use the references to access important values if needed for this question.
how many grams of pbbr₂ will precipitate when excess mnbr₂ solution is added to 47.0 ml of 0.757 m pb(no₃)₂ solution?
pb(no₃)₂(aq) + mnbr₂(aq) → pbbr₂(s) + mn(no₃)₂(aq)
g

Explanation:

Step1: Calculate the moles of \(Pb(NO_3)_2\)

Use the formula \(n = C\times V\). Given \(C = 0.757\space M\) and \(V=47.0\space mL=0.0470\space L\).
\(n_{Pb(NO_3)_2}=0.757\space M\times0.0470\space L = 0.035579\space mol\)

Step2: Determine the moles of \(PbBr_2\)

From the balanced chemical equation \(Pb(NO_3)_2(aq)+MnBr_2(aq)\to PbBr_2(s)+Mn(NO_3)_2(aq)\), the mole ratio of \(Pb(NO_3)_2\) to \(PbBr_2\) is \(1:1\). So \(n_{PbBr_2}=n_{Pb(NO_3)_2} = 0.035579\space mol\)

Step3: Calculate the mass of \(PbBr_2\)

The molar mass of \(PbBr_2\) is \(M=(207.2 + 2\times79.904)\space g/mol= 367.008\space g/mol\). Use the formula \(m = n\times M\)
\(m_{PbBr_2}=0.035579\space mol\times367.008\space g/mol\approx12.96\space g\)

Answer:

\(13.0\space g\) (rounded to three significant figures)