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 mass of precipitate that can form?(ans.4.61g)
3. a student mixes 15.0ml of 0.250mol/l aqueous sodium hydroxide with 20.0ml of 0.400mol/l aqueous aluminum nitrate.

a) write the chemical equation for the reaction.
b) calculate the maximum mass of precipitate that forms.(ans.0.0975g)

Explanation:

Part (a)

The reaction is between sodium hydroxide ($\ce{NaOH}$) and aluminum nitrate ($\ce{Al(NO3)3}$). The products are aluminum hydroxide precipitate ($\ce{Al(OH)3}$) and sodium nitrate ($\ce{NaNO3}$). Balancing the equation: 3 moles of $\ce{NaOH}$ react with 1 mole of $\ce{Al(NO3)3}$ to form 1 mole of $\ce{Al(OH)3}$ and 3 moles of $\ce{NaNO3}$.

Step1: Calculate moles of reactants

Moles of $\ce{NaOH}$: $n(\ce{NaOH}) = c \times V = 0.250\,\text{mol/L} \times 0.0150\,\text{L} = 0.00375\,\text{mol}$
Moles of $\ce{Al(NO3)3}$: $n(\ce{Al(NO3)3}) = 0.400\,\text{mol/L} \times 0.0200\,\text{L} = 0.00800\,\text{mol}$

Step2: Determine limiting reactant

From the equation, 3 mol $\ce{NaOH}$ reacts with 1 mol $\ce{Al(NO3)3}$.
Moles of $\ce{Al(NO3)3}$ required for 0.00375 mol $\ce{NaOH}$: $\frac{0.00375}{3} = 0.00125\,\text{mol}$
Since 0.00800 mol $\ce{Al(NO3)3}$ is present (more than 0.00125 mol), $\ce{NaOH}$ is limiting.

Step3: Moles of $\ce{Al(OH)3}$ formed

From the equation, 3 mol $\ce{NaOH}$ produces 1 mol $\ce{Al(OH)3}$.
Moles of $\ce{Al(OH)3}$: $\frac{0.00375}{3} = 0.00125\,\text{mol}$

Step4: Calculate mass of $\ce{Al(OH)3}$

Molar mass of $\ce{Al(OH)3}$: $27 + 3(16 + 1) = 78\,\text{g/mol}$
Mass: $m = n \times M = 0.00125\,\text{mol} \times 78\,\text{g/mol} = 0.0975\,\text{g}$ (matches the given answer, slight rounding differences possible)

Answer:

$\ce{Al(NO3)3 + 3NaOH -> Al(OH)3(s) + 3NaNO3}$

Part (b)