When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4.

/ Exams & Certification / By

When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4. The balanced equation is: 2AgNO3(aq) + Na2SO4(aq) → Ag2SO4(s) + 2NaNO3(aq) How many grams of silver sulfate are expected when a solution containing 0.50 mol AgNO3 is mixed with a solution containing 0.30 mol Na2SO4

The Correct Answer and Explanation is:

To calculate the amount of silver sulfate (Ag2SO4) formed when mixing solutions of silver nitrate (AgNO3) and sodium sulfate (Na2SO4), follow these steps:

Step 1: Write the balanced equation

The balanced chemical equation is:2AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+2NaNO3(aq)2AgNO3(aq) + Na2SO4(aq) \rightarrow Ag2SO4(s) + 2NaNO3(aq)2AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+2NaNO3(aq)

Step 2: Determine the limiting reagent

First, let’s look at the stoichiometric coefficients of the equation:

  • 2 moles of AgNO3 react with 1 mole of Na2SO4 to produce 1 mole of Ag2SO4.

You are given:

  • 0.50 mol of AgNO3
  • 0.30 mol of Na2SO4

Now, compare the mole ratios:

  • According to the equation, 2 moles of AgNO3 need 1 mole of Na2SO4.
  • To fully react with 0.50 mol of AgNO3, we would need 0.502=0.25\frac{0.50}{2} = 0.2520.50​=0.25 mol of Na2SO4.

Since we have 0.30 mol of Na2SO4 available, and only 0.25 mol is needed, Na2SO4 is in excess, and AgNO3 is the limiting reagent.

Step 3: Calculate the amount of Ag2SO4 formed

Since AgNO3 is the limiting reagent, use its amount to calculate the amount of Ag2SO4:

  • From the balanced equation, 2 moles of AgNO3 produce 1 mole of Ag2SO4.
  • Therefore, 0.50 mol of AgNO3 will produce 0.502=0.25\frac{0.50}{2} = 0.2520.50​=0.25 mol of Ag2SO4.

Step 4: Convert moles of Ag2SO4 to grams

Now, calculate the mass of Ag2SO4 formed. The molar mass of Ag2SO4 is:Molar mass of Ag2SO4=2(107.87 g/mol)+32.07 g/mol+4(16.00 g/mol)=311.87 g/mol\text{Molar mass of Ag2SO4} = 2(107.87 \, \text{g/mol}) + 32.07 \, \text{g/mol} + 4(16.00 \, \text{g/mol}) = 311.87 \, \text{g/mol}Molar mass of Ag2SO4=2(107.87g/mol)+32.07g/mol+4(16.00g/mol)=311.87g/mol

Thus, the mass of Ag2SO4 formed is:Mass of Ag2SO4=0.25 mol×311.87 g/mol=77.97 g\text{Mass of Ag2SO4} = 0.25 \, \text{mol} \times 311.87 \, \text{g/mol} = 77.97 \, \text{g}Mass of Ag2SO4=0.25mol×311.87g/mol=77.97g

Final Answer:

The mass of silver sulfate (Ag2SO4) expected to form is 77.97 g.

Related Posts

Scroll to Top