25.0 mL of 0.125 M silver nitrate are mixed with 35.0mL of 0.100 M sodium sulfate to give solid silver sulfate. Write the net ionic equation, and determine the mass in grams of silver sulfate produced.
I think the net ionic equation is
2Ag(aq) + SO4^-2(aq) ---> AgSO4(s)
but not sure what to do next...
M=mol/L
"25.0 mL of 0.125 M silver nitrate" = (25/1000 L)(0.125 mol/L)
"35.0mL of 0.100 M sodium sulfate"=
(35/1000 L)(0.100 mol/L)
now that you have the moles of Ag and SO4^2- convert to moles of AgSO4 through the mole ratio and smaller value is the limiting reagent
take the limiting reagent and multiply by the molar mass AgSO4
I think silver sulfate is Ag2SO4.
Also I think this problem fosters the idea that Ag2SO4 is relatively insoluble and it isn't. In fact, about 0.250 g of Ag2SO4 will dissolve in this solution and that amount will be lost.
you're right doc its Ag2SO4, thanks guys!
To determine the mass of silver sulfate produced, you need to use stoichiometry and the given information about the amounts of silver nitrate and sodium sulfate used.
The balanced net ionic equation you wrote is correct:
2Ag+(aq) + SO4^2-(aq) → Ag2SO4(s)
From the balanced equation, we can see that 2 moles of silver ions (Ag+) react with 1 mole of sulfate ions (SO4^2-) to produce 1 mole of solid silver sulfate (Ag2SO4).
Step 1: Calculate the number of moles of silver nitrate used.
Given:
Volume of silver nitrate solution (V1) = 25.0 mL = 0.025 L
Concentration of silver nitrate (C1) = 0.125 M
Use the formula:
Moles = Volume (in liters) × Concentration
Moles of silver nitrate (n1) = V1 × C1 = 0.025 L × 0.125 M
Step 2: Calculate the number of moles of sodium sulfate used.
Given:
Volume of sodium sulfate solution (V2) = 35.0 mL = 0.035 L
Concentration of sodium sulfate (C2) = 0.100 M
Moles of sodium sulfate (n2) = V2 × C2 = 0.035 L × 0.100 M
Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the ratios of the moles of reactants:
the molar ratio from the balanced equation is 2:1 for Ag+:SO4^2- ions.
Since the ratio of moles of Ag+:SO4^2- is 2:1, the limiting reactant will be the one that produces fewer moles of Ag2SO4.
Let's compare the moles of silver nitrate (n1) and sodium sulfate (n2).
If n1 ≤ 2n2, then silver nitrate is the limiting reactant.
If n1 > 2n2, then sodium sulfate is the limiting reactant.
Step 4: Calculate the moles of Ag2SO4 formed.
Depending on the limiting reactant, you can proceed with the appropriate value.
If silver nitrate is the limiting reactant (n1 ≤ 2n2):
Moles of Ag2SO4 formed = n1 / 2
If sodium sulfate is the limiting reactant (n1 > 2n2):
Moles of Ag2SO4 formed = n2
Step 5: Calculate the mass of Ag2SO4 formed.
To calculate the mass of Ag2SO4, you need to know the molar mass of Ag2SO4.
The molar mass of Ag2SO4 = (2 × atomic mass of Ag) + atomic mass of S + (4 × atomic mass of O)
Now, multiply the moles of Ag2SO4 formed by the molar mass of Ag2SO4 to get the mass (using the appropriate value from step 4).
Mass of Ag2SO4 formed = Moles of Ag2SO4 formed × Molar mass of Ag2SO4
Note: Make sure to convert the volume from milliliters to liters for all calculations.
Following these steps should give you the final mass of silver sulfate produced.