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

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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?
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 solve this problem, we need to determine the amount of silver sulfate (Ag2SO4) produced in the reaction when mixing 0.50 mol of AgNO3 and 0.30 mol of Na2SO4. We’ll do this step by step, considering the stoichiometry of the reaction.

Step 1: Write the balanced chemical equation

The balanced equation for the reaction is:2AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+2NaNO3(aq)2AgNO_3(aq) + Na_2SO_4(aq) \rightarrow Ag_2SO_4(s) + 2NaNO_3(aq)2AgNO3​(aq)+Na2​SO4​(aq)→Ag2​SO4​(s)+2NaNO3​(aq)

From this equation, we see that:

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

Step 2: Identify the limiting reactant

We are given:

  • 0.50 mol of AgNO3
  • 0.30 mol of Na2SO4

From the equation, we need 2 moles of AgNO3 for every 1 mole of Na2SO4. Let’s check how much AgNO3 is needed for 0.30 mol of Na2SO4:0.30 mol Na2SO4×2 mol AgNO31 mol Na2SO4=0.60 mol AgNO30.30 \, \text{mol Na}_2\text{SO}_4 \times \frac{2 \, \text{mol AgNO}_3}{1 \, \text{mol Na}_2\text{SO}_4} = 0.60 \, \text{mol AgNO}_30.30mol Na2​SO4​×1mol Na2​SO4​2mol AgNO3​​=0.60mol AgNO3​

We have 0.50 mol of AgNO3, but we need 0.60 mol to completely react with 0.30 mol of Na2SO4. Therefore, AgNO3 is the limiting reactant.

Step 3: Calculate the amount of silver sulfate produced

From the balanced equation, 2 moles of AgNO3 produce 1 mole of Ag2SO4. Since AgNO3 is limiting, we can calculate how many moles of Ag2SO4 are produced from 0.50 mol of AgNO3:0.50 mol AgNO3×1 mol Ag2SO42 mol AgNO3=0.25 mol Ag2SO40.50 \, \text{mol AgNO}_3 \times \frac{1 \, \text{mol Ag}_2\text{SO}_4}{2 \, \text{mol AgNO}_3} = 0.25 \, \text{mol Ag}_2\text{SO}_40.50mol AgNO3​×2mol AgNO3​1mol Ag2​SO4​​=0.25mol Ag2​SO4​

Step 4: Convert moles of Ag2SO4 to grams

Next, we need to convert the moles of Ag2SO4 to grams. The molar mass of Ag2SO4 is calculated as follows:

  • Silver (Ag): 107.87 g/mol107.87 \, \text{g/mol}107.87g/mol
  • Sulfur (S): 32.07 g/mol32.07 \, \text{g/mol}32.07g/mol
  • Oxygen (O): 16.00 g/mol16.00 \, \text{g/mol}16.00g/mol

The molar mass of Ag2SO4 is:(2×107.87)+32.07+(4×16.00)=215.74+32.07+64.00=311.81 g/mol(2 \times 107.87) + 32.07 + (4 \times 16.00) = 215.74 + 32.07 + 64.00 = 311.81 \, \text{g/mol}(2×107.87)+32.07+(4×16.00)=215.74+32.07+64.00=311.81g/mol

Now, calculate the mass of Ag2SO4 produced:0.25 mol Ag2SO4×311.81 g/mol=77.95 g Ag2SO40.25 \, \text{mol Ag}_2\text{SO}_4 \times 311.81 \, \text{g/mol} = 77.95 \, \text{g Ag}_2\text{SO}_40.25mol Ag2​SO4​×311.81g/mol=77.95g Ag2​SO4​

Final Answer:

The mass of silver sulfate (Ag2SO4) produced is 77.95 grams.

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