Given the following reaction

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Given the following reaction: (hint: balance the reaction first) Na2SO4+ AgNO3→ + Ag2SO4 If you start with 60 g of a 3.0 M Na2SO4 solution and 40 g of a 3.0 M AgNO3 solution, a) determine the limiting reagent b) determine the number of grams of excess reagent left c)determine the number of moles of NaNO3 produced. c) determine the number of moles of Ag2SO4 produced

The Correct Answer and Explanation is:

To solve this problem, we need to go through a few steps systematically:

Step 1: Write the balanced chemical equation

The reaction is:Na2SO4(aq)+2AgNO3(aq)→Ag2SO4(s)+2NaNO3(aq)\text{Na}_2\text{SO}_4 (aq) + 2\text{AgNO}_3 (aq) \rightarrow \text{Ag}_2\text{SO}_4 (s) + 2\text{NaNO}_3 (aq)Na2​SO4​(aq)+2AgNO3​(aq)→Ag2​SO4​(s)+2NaNO3​(aq)

This balanced equation tells us that 1 mole of Na2SO4 reacts with 2 moles of AgNO3 to produce 1 mole of Ag2SO4 and 2 moles of NaNO3.

Step 2: Calculate moles of Na2SO4 and AgNO3

We are given:

  • 60 g of Na2SO4 (3.0 M solution)
  • 40 g of AgNO3 (3.0 M solution)

Moles of Na2SO4:

First, calculate the molar mass of Na2SO4:Na2SO4=2×22.99(Na)+32.07(S)+4×16.00(O)=142.05 g/mol\text{Na}_2\text{SO}_4 = 2 \times 22.99 (\text{Na}) + 32.07 (\text{S}) + 4 \times 16.00 (\text{O}) = 142.05 \, \text{g/mol}Na2​SO4​=2×22.99(Na)+32.07(S)+4×16.00(O)=142.05g/mol

Now, calculate moles of Na2SO4:moles of Na2SO4=60 g142.05 g/mol=0.422 mol\text{moles of Na}_2\text{SO}_4 = \frac{60 \, \text{g}}{142.05 \, \text{g/mol}} = 0.422 \, \text{mol}moles of Na2​SO4​=142.05g/mol60g​=0.422mol

Moles of AgNO3:

The molar mass of AgNO3 is:AgNO3=107.87(Ag)+14.01(N)+3×16.00(O)=169.87 g/mol\text{AgNO}_3 = 107.87 (\text{Ag}) + 14.01 (\text{N}) + 3 \times 16.00 (\text{O}) = 169.87 \, \text{g/mol}AgNO3​=107.87(Ag)+14.01(N)+3×16.00(O)=169.87g/mol

Now, calculate moles of AgNO3:moles of AgNO3=40 g169.87 g/mol=0.235 mol\text{moles of AgNO}_3 = \frac{40 \, \text{g}}{169.87 \, \text{g/mol}} = 0.235 \, \text{mol}moles of AgNO3​=169.87g/mol40g​=0.235mol

Step 3: Determine the limiting reagent

From the balanced equation, we see that 1 mole of Na2SO4 reacts with 2 moles of AgNO3. So, for 0.422 moles of Na2SO4, we need:0.422 mol Na2SO4×2 mol AgNO31 mol Na2SO4=0.844 mol AgNO30.422 \, \text{mol Na}_2\text{SO}_4 \times \frac{2 \, \text{mol AgNO}_3}{1 \, \text{mol Na}_2\text{SO}_4} = 0.844 \, \text{mol AgNO}_30.422mol Na2​SO4​×1mol Na2​SO4​2mol AgNO3​​=0.844mol AgNO3​

However, we only have 0.235 moles of AgNO3, which is less than the required 0.844 moles. Therefore, AgNO3 is the limiting reagent.

Step 4: Determine the amount of excess reagent (Na2SO4) left

From the balanced equation, 2 moles of AgNO3 react with 1 mole of Na2SO4. Since we have 0.235 moles of AgNO3, we need:0.235 mol AgNO3×1 mol Na2SO42 mol AgNO3=0.1175 mol Na2SO40.235 \, \text{mol AgNO}_3 \times \frac{1 \, \text{mol Na}_2\text{SO}_4}{2 \, \text{mol AgNO}_3} = 0.1175 \, \text{mol Na}_2\text{SO}_40.235mol AgNO3​×2mol AgNO3​1mol Na2​SO4​​=0.1175mol Na2​SO4​

This means 0.1175 moles of Na2SO4 will react. Initially, we had 0.422 moles of Na2SO4, so the amount of Na2SO4 remaining is:0.422 mol−0.1175 mol=0.3045 mol Na2SO40.422 \, \text{mol} – 0.1175 \, \text{mol} = 0.3045 \, \text{mol Na}_2\text{SO}_40.422mol−0.1175mol=0.3045mol Na2​SO4​

Now, convert the moles of Na2SO4 remaining back to grams:0.3045 mol×142.05 g/mol=43.2 g Na2SO40.3045 \, \text{mol} \times 142.05 \, \text{g/mol} = 43.2 \, \text{g Na}_2\text{SO}_40.3045mol×142.05g/mol=43.2g Na2​SO4​

So, 43.2 g of Na2SO4 remain as excess.

Step 5: Determine the number of moles of NaNO3 produced

From the balanced equation, 2 moles of AgNO3 produce 2 moles of NaNO3. Since we are using 0.235 moles of AgNO3, the moles of NaNO3 produced are:0.235 mol AgNO3×2 mol NaNO32 mol AgNO3=0.235 mol NaNO30.235 \, \text{mol AgNO}_3 \times \frac{2 \, \text{mol NaNO}_3}{2 \, \text{mol AgNO}_3} = 0.235 \, \text{mol NaNO}_30.235mol AgNO3​×2mol AgNO3​2mol NaNO3​​=0.235mol NaNO3​

So, 0.235 moles of NaNO3 will be produced.

Step 6: Determine the number of moles of Ag2SO4 produced

From the balanced equation, 2 moles of AgNO3 produce 1 mole of Ag2SO4. Therefore, the moles of Ag2SO4 produced are:0.235 mol AgNO3×1 mol Ag2SO42 mol AgNO3=0.1175 mol Ag2SO40.235 \, \text{mol AgNO}_3 \times \frac{1 \, \text{mol Ag}_2\text{SO}_4}{2 \, \text{mol AgNO}_3} = 0.1175 \, \text{mol Ag}_2\text{SO}_40.235mol AgNO3​×2mol AgNO3​1mol Ag2​SO4​​=0.1175mol Ag2​SO4​

So, 0.1175 moles of Ag2SO4 will be produced.

Final Answers:

  • a) The limiting reagent is AgNO3.
  • b) The number of grams of excess reagent (Na2SO4) left is 43.2 g.
  • c) The number of moles of NaNO3 produced is 0.235 mol.
  • d) The number of moles of Ag2SO4 produced is 0.1175 mol.

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