Question

What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)

Short answer

The mass of solid aluminum hydroxide produced when $50.0 \mathrm{~mL}$ of $0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$ is added to $200.0 \mathrm{~mL}$ of $0.100 \mathrm{M} \mathrm{KOH}$ is approximately \(0.780~g\).

Step-by-step solution

1. Write the balanced chemical equation

The reaction between aluminum nitrate and potassium hydroxide produces solid aluminum hydroxide and potassium nitrate: \[ Al(NO_3)_3 + 3KOH \rightarrow Al(OH)_3(s) + 3KNO_3 \]

2. Calculate the moles of Al(NO3)3 and KOH

We'll calculate the moles of Al(NO_3)_3 and KOH using their volume and concentration, using the formula: \[ moles = volume (L) \times concentration (M) \] For Al(NO_3)_3: \(moles = 50.0 \times 10^{-3} L \times 0.200 M = 0.0100 mol\) For KOH: \(moles = 200.0 \times 10^{-3} L \times 0.100 M = 0.0200 mol\)

3. Determine the limiting reactant

To determine the limiting reactant, we'll compare the mole ratios of Al(NO_3)_3 and KOH in the balanced chemical equation. According to the equation, we need 1 mole of Al(NO_3)_3 to react with 3 moles of KOH. Mole ratio = moles of KOH / moles of Al(NO_3)_3 Mole ratio = \(0.0200 / 0.0100 = 2.00\) Since the mole ratio is less than the required 3:1, KOH is the limiting reactant.

4. Calculate the theoretical yield of Al(OH)3

We can use the stoichiometry from the balanced chemical equation to find the theoretical yield of Al(OH)_3: 1 mol of Al(NO_3)_3 reacts with 3 mol of KOH to produce 1 mol of Al(OH)_3 The moles of Al(OH)_3 produced will be equal to the moles of Al(NO_3)_3 available: \(moles \space of \space Al(OH)_3 = 0.0100 \space mol\)

5. Convert moles of Al(OH)3 to mass

We'll now convert the moles of aluminum hydroxide to its mass using its molar mass. The molar mass of Al(OH)_3 is approximately: \(1 \times Al \approx 26.98~g/mol\) \(3 \times O \approx 47.997~g/mol\) \(3 \times H \approx 3.024~g/mol\) Molar mass of Al(OH)_3 ≈ \(26.98 + 47.997 + 3.024 \approx 78.001~g/mol\) Mass of Al(OH)_3 = moles × molar mass Mass of Al(OH)_3 ≈ \(0.0100~mol \times 78.001~g/mol = 0.780~g\) So, the mass of solid aluminum hydroxide produced in the reaction is approximately \(0.780~g\).