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 determine the grams of silver sulfate (Ag2SO4) produced when solutions of silver nitrate (AgNO3) and sodium sulfate (Na2SO4) are mixed, follow these steps:
1. Write and Balance the Equation:
The balanced equation is:2AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+2NaNO3(aq)2AgNO3(aq) + Na2SO4(aq) → Ag2SO4(s) + 2NaNO3(aq)2AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+2NaNO3(aq)
2. Identify the Limiting Reagent:
From the balanced equation, 2 moles of AgNO3 react with 1 mole of Na2SO4 to form 1 mole of Ag2SO4.
Given:
- 0.50 mol of AgNO3
- 0.30 mol of Na2SO4
From the equation:
- 2 moles of AgNO3 need 1 mole of Na2SO4.
- For 0.50 moles of AgNO3, you would need:
Na2SO4 needed=0.50 mol AgNO32=0.25 mol Na2SO4\text{Na2SO4 needed} = \frac{0.50 \, \text{mol} \, \text{AgNO3}}{2} = 0.25 \, \text{mol} \, \text{Na2SO4}Na2SO4 needed=20.50molAgNO3=0.25molNa2SO4
Since you have 0.30 mol Na2SO4 available, which is more than the required 0.25 mol, AgNO3 is the limiting reagent.
3. Determine the Moles of Silver Sulfate Produced:
From the balanced equation:
- 2 moles of AgNO3 produce 1 mole of Ag2SO4.
- Since 0.50 moles of AgNO3 are used, the moles of Ag2SO4 produced are:
Moles of Ag2SO4=0.50 mol AgNO32=0.25 mol Ag2SO4\text{Moles of Ag2SO4} = \frac{0.50 \, \text{mol AgNO3}}{2} = 0.25 \, \text{mol Ag2SO4}Moles of Ag2SO4=20.50mol AgNO3=0.25mol Ag2SO4
4. Convert Moles of Ag2SO4 to Grams:
The molar mass of Ag2SO4 is calculated as:Molar mass of Ag2SO4=(2×107.87)+32.07+(4×16.00)=311.87 g/mol\text{Molar mass of Ag2SO4} = (2 \times 107.87) + 32.07 + (4 \times 16.00) = 311.87 \, \text{g/mol}Molar mass of Ag2SO4=(2×107.87)+32.07+(4×16.00)=311.87g/mol
Now, convert the moles of Ag2SO4 to grams: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) produced is approximately 77.97 grams.
Explanation:
Since AgNO3 is the limiting reagent, the amount of silver sulfate produced depends directly on the amount of AgNO3. Using stoichiometric relationships, the calculated moles of Ag2SO4 are converted to grams, giving the final yield of 77.97 grams.
