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:
Let’s break this problem down step by step:
Step 1: Write the balanced chemical equation.
The unbalanced equation is:
Na2SO4(aq)+AgNO3(aq)→Ag2SO4(s)+NaNO3(aq)\text{Na}_2\text{SO}_4 (aq) + \text{AgNO}_3 (aq) \rightarrow \text{Ag}_2\text{SO}_4 (s) + \text{NaNO}_3 (aq)Na2SO4(aq)+AgNO3(aq)→Ag2SO4(s)+NaNO3(aq)
Now, balance the equation:
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)Na2SO4(aq)+2AgNO3(aq)→Ag2SO4(s)+2NaNO3(aq)
Step 2: Find the moles of each reactant.
- Na2SO4:
Given: 60 g of Na2SO4 and a concentration of 3.0 M.
Molar mass of Na2SO4 = 2(22.99)+32.06+4(16.00)=142.04 g/mol2(22.99) + 32.06 + 4(16.00) = 142.04 \, \text{g/mol}2(22.99)+32.06+4(16.00)=142.04g/mol.
To calculate the moles of Na2SO4:moles of Na2SO4=massmolar mass=60 g142.04 g/mol=0.4226 mol\text{moles of Na2SO4} = \frac{\text{mass}}{\text{molar mass}} = \frac{60 \, \text{g}}{142.04 \, \text{g/mol}} = 0.4226 \, \text{mol}moles of Na2SO4=molar massmass=142.04g/mol60g=0.4226mol
- AgNO3:
Given: 40 g of AgNO3 and a concentration of 3.0 M.
Molar mass of AgNO3 = 107.87+14.01+3(16.00)=169.87 g/mol107.87 + 14.01 + 3(16.00) = 169.87 \, \text{g/mol}107.87+14.01+3(16.00)=169.87g/mol.
To calculate the moles of AgNO3:moles of AgNO3=massmolar mass=40 g169.87 g/mol=0.2358 mol\text{moles of AgNO3} = \frac{\text{mass}}{\text{molar mass}} = \frac{40 \, \text{g}}{169.87 \, \text{g/mol}} = 0.2358 \, \text{mol}moles of AgNO3=molar massmass=169.87g/mol40g=0.2358mol
Step 3: Determine the limiting reagent.
The balanced equation shows that 1 mole of Na2SO4 reacts with 2 moles of AgNO3. Therefore, for 0.4226 moles of Na2SO4, we need:0.4226 mol Na2SO4×2=0.8452 mol AgNO30.4226 \, \text{mol Na2SO4} \times 2 = 0.8452 \, \text{mol AgNO3}0.4226mol Na2SO4×2=0.8452mol AgNO3
However, we only have 0.2358 moles of AgNO3, which is less than the 0.8452 moles required. Thus, AgNO3 is the limiting reagent.
Step 4: Find the amount of excess reagent left (Na2SO4).
From the balanced equation, 2 moles of AgNO3 react with 1 mole of Na2SO4. Since we have 0.2358 moles of AgNO3, it will react with:0.2358 mol AgNO32=0.1179 mol Na2SO4\frac{0.2358 \, \text{mol AgNO3}}{2} = 0.1179 \, \text{mol Na2SO4}20.2358mol AgNO3=0.1179mol Na2SO4
The initial moles of Na2SO4 were 0.4226. So, the moles of Na2SO4 left over are:0.4226 mol Na2SO4−0.1179 mol Na2SO4=0.3047 mol Na2SO40.4226 \, \text{mol Na2SO4} – 0.1179 \, \text{mol Na2SO4} = 0.3047 \, \text{mol Na2SO4}0.4226mol Na2SO4−0.1179mol Na2SO4=0.3047mol Na2SO4
Now, convert the excess moles of Na2SO4 to grams:mass of Na2SO4 left=0.3047 mol×142.04 g/mol=43.2 g\text{mass of Na2SO4 left} = 0.3047 \, \text{mol} \times 142.04 \, \text{g/mol} = 43.2 \, \text{g}mass of Na2SO4 left=0.3047mol×142.04g/mol=43.2g
Step 5: Find the moles of NaNO3 produced.
From the balanced equation, for every 2 moles of AgNO3, 2 moles of NaNO3 are produced. Since we have 0.2358 moles of AgNO3, the moles of NaNO3 produced are:0.2358 mol NaNO30.2358 \, \text{mol NaNO3}0.2358mol NaNO3
Step 6: Find the moles of Ag2SO4 produced.
Again, from the balanced equation, 2 moles of AgNO3 produce 1 mole of Ag2SO4. Therefore, the moles of Ag2SO4 produced are:0.2358 mol AgNO32=0.1179 mol Ag2SO4\frac{0.2358 \, \text{mol AgNO3}}{2} = 0.1179 \, \text{mol Ag2SO4}20.2358mol AgNO3=0.1179mol Ag2SO4
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 0.2358 mol.
d) The number of moles of Ag2SO4 produced is 0.1179 mol.
