Question

1. 8.76 g of sodium sulfide is added to 350 ml of 0.250 mol/l lead(ii) nitrate solution. calculate the maximum mass of precipitate that can form. (ans.20.9g)
2. 25.0ml of 0.400mol/l pb(no₃)₂ is mixed with 300ml of 0.220 mol/l ki. what is the maximum

Explanation:

Step1: Write the balanced reaction

The reaction between sodium sulfide ($\ce{Na_{2}S}$) and lead(II) nitrate ($\ce{Pb(NO_{3})_{2}}$) is:
$$\ce{Na_{2}S(aq) + Pb(NO_{3})_{2}(aq) -> PbS(s) + 2NaNO_{3}(aq)}$$
The precipitate is $\ce{PbS}$.

Step2: Calculate moles of reactants

• Moles of $\ce{Na_{2}S}$:

Mass of $\ce{Na_{2}S}$ = 8.76 g, molar mass of $\ce{Na_{2}S}$ = $2\times23 + 32 = 78$ g/mol.
Moles of $\ce{Na_{2}S}$ = $\frac{8.76\ \text{g}}{78\ \text{g/mol}} = 0.1123\ \text{mol}$.

• Moles of $\ce{Pb(NO_{3})_{2}}$:

Volume = 350 mL = 0.350 L, concentration = 0.250 mol/L.
Moles of $\ce{Pb(NO_{3})_{2}}$ = $0.250\ \text{mol/L} \times 0.350\ \text{L} = 0.0875\ \text{mol}$.

Step3: Determine limiting reactant

From the balanced equation, 1 mol $\ce{Na_{2}S}$ reacts with 1 mol $\ce{Pb(NO_{3})_{2}}$.

• Moles of $\ce{Na_{2}S}$ available: 0.1123 mol
• Moles of $\ce{Pb(NO_{3})_{2}}$ available: 0.0875 mol

Since $\ce{Pb(NO_{3})_{2}}$ has fewer moles, it is the limiting reactant.

Step4: Moles of $\ce{PbS}$ formed

From the equation, 1 mol $\ce{Pb(NO_{3})_{2}}$ produces 1 mol $\ce{PbS}$.
Thus, moles of $\ce{PbS}$ = moles of limiting reactant = 0.0875 mol.

Step5: Calculate mass of $\ce{PbS}$

Molar mass of $\ce{PbS}$ = $207.2 + 32.07 = 239.27$ g/mol.
Mass of $\ce{PbS}$ = moles × molar mass = $0.0875\ \text{mol} \times 239.27\ \text{g/mol} \approx 20.9\ \text{g}$.

Answer:

20.9 g