Question

ACS Exams Study Question - Chapter 5 (Solutions and Aqueous Reactions) 13A. Aluminum nitrate (Al (N*O_{3}) 3 ) reacts with sodium sulfide (Na, S) to form a precipitate: 2Al(NO3)3(aq) + 3Na₂S(aq) Al , overline S 3 (s) + NaNO3(aq) molar mass/g-mat- Al₂S 150.17 0251 Ma m/L When reacting 25.0 mL of the aluminum nitrate solution with an excess of sodium sulfide, 1.085 g of the precipitate is recovered at a 91.8% yield. What was the molar concentration of the aluminum nitrate solution?

Answer 1

To find the molar concentration of the aluminum nitrate solution, follow a series of steps using the given information.

Step-by-step explanation:

To find the molar concentration of the aluminum nitrate solution, we need to use the given information and follow a series of steps:

1. Convert the mass of the precipitate (Al2S3) to moles.
2. Use the balanced equation to determine the mole ratio between Al(NO3)3 and Al2S3.
3. Convert the moles of Al(NO3)3 to find the initial moles in the solution.
4. Divide the moles by the volume of the solution (25.0 mL) to find the molar concentration.

Using these steps, the molar concentration of the aluminum nitrate solution can be calculated.

Answer 2

To calculate the molar concentration of the aluminum nitrate solution, you need to use the given information about the mass of the precipitate and the percent yield. The molar concentration is 0.576 M.

Step-by-step explanation:

To calculate the molar concentration of the aluminum nitrate solution, we can use the given information about the mass of the precipitate and the percent yield.

First, calculate the moles of the precipitate by dividing the mass by the molar mass of Al₂S.

Then, calculate the moles of Al(NO3)3 in the reactants using the stoichiometry of the balanced equation.

Finally, divide the moles of Al(NO3)3 by the volume of the solution in liters to find the molar concentration.

Example:

Mass of Al₂S precipitate = 1.085 g

Percent yield = 91.8%

Molar mass of Al₂S = 150.17 g/mol

Volume of solution = 25.0 mL

= 0.025 L

Step 1: Calculate moles of Al₂S precipitate:

Moles = mass / molar mass

= 1.085 g / 150.17 g/mol

= 0.00722 mol

Step 2: Calculate moles of Al(NO3)3 in the reactants:

From the balanced equation, the stoichiometry is 2:1 for Al(NO3)3:Al₂S

Moles of Al(NO3)3 = 2 * 0.00722 mol

= 0.0144 mol

Step 3: Calculate molar concentration:

Molar concentration = moles / volume

= 0.0144 mol / 0.025 L

= 0.576 M