Question

final exam 2025
solutions:
use the following information to answer the next question.
a 150.0 ml solution of lead(ii) nitrate with a concentration of 0.750 mol/l is added to
an excess of potassium iodide.

1. calculate the mass of the precipitate produced. (show all work to get full marks)

your answer:

Explanation:

Step1: Write the balanced chemical equation

$$\text{Pb(NO}_3\text{)}_2 + 2\text{KI} ightarrow\text{PbI}_2\downarrow+ 2\text{KNO}_3$$

Step2: Calculate the moles of lead(II) nitrate

Molarity formula $n = C\times V$. Given $C = 0.750\space mol/L$ and $V=150.0\space mL=0.150\space L$.
$$n(\text{Pb(NO}_3\text{)}_2)=0.750\space mol/L\times0.150\space L = 0.1125\space mol$$

Step3: Determine the moles of precipitate ($\text{PbI}_2$)

From the balanced equation, mole ratio $\text{Pb(NO}_3\text{)}_2:\text{PbI}_2 = 1:1$. So $n(\text{PbI}_2)=n(\text{Pb(NO}_3\text{)}_2) = 0.1125\space mol$

Step4: Calculate the molar mass of $\text{PbI}_2$

Molar mass of $\text{Pb}=207.2\space g/mol$, molar mass of $\text{I}=126.9\space g/mol$.
$M(\text{PbI}_2)=207.2+(2\times126.9)= 207.2 + 253.8=461\space g/mol$

Step5: Calculate the mass of $\text{PbI}_2$

Mass formula $m=n\times M$.
$$m(\text{PbI}_2)=0.1125\space mol\times461\space g/mol = 51.9\space g$$

Answer:

$51.9\space g$