If you react 2.0 g AgNO3 and 2.5 g Na2SO4, what is the percent yield if the actual yield is 1.6 g Ag2SO4? 2AgNO3 + Na2SO4 → Ag2SO4 + 2NaNO3
Do you know the limiting reagent?
1. Convert g AgNO3 to moles. moles = grams/molar mass.
2. Convert g Na2SO4 to moles.
3a. Using the coefficients in the balanced equation, convert moles AgNO3 to moles Ag2SO4.
3b. Same process convert moles Na2SO4 to moles Ag2SO4.
3c. Both answers can't be correct; the correct one in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
4. The smaller value moles x molar mass = grams Ag2SO4. This is the theoretical yield which we can call TE.
5. %yield = (actual yield/TE)*100 = ?
12312
To calculate the percent yield, you need to compare the actual yield with the theoretical yield. The theoretical yield is the maximum amount of product that can be formed based on stoichiometry.
Step 1: Calculate the molar masses of the reactants and product:
- AgNO3: 107.87 g/mol (Ag: 107.87 g/mol, N: 14.01 g/mol, O: 16.00 g/mol)
- Na2SO4: 142.04 g/mol (Na: 22.99 g/mol, S: 32.06 g/mol, O: 16.00 g/mol)
- Ag2SO4: 311.81 g/mol (Ag: 107.87 g/mol, S: 32.06 g/mol, O: 16.00 g/mol)
Step 2: Calculate the number of moles of reactants using their masses and molar masses:
- AgNO3: 2.0 g ÷ 107.87 g/mol = 0.0185 mol
- Na2SO4: 2.5 g ÷ 142.04 g/mol = 0.0176 mol
Step 3: Determine the limiting reactant (the one that is completely consumed). This can be done by comparing the mole ratios of the reactants based on the balanced equation:
- From the balanced equation, the ratio of AgNO3 to Na2SO4 is 2:1.
- Since AgNO3 and Na2SO4 have equal moles (0.0185 mol and 0.0176 mol, respectively), the limiting reactant is Na2SO4.
Step 4: Calculate the theoretical yield of Ag2SO4 using the mole ratio from the balanced equation:
- From the balanced equation, the ratio of Ag2SO4 to Na2SO4 is 1:1.
- Therefore, the theoretical yield of Ag2SO4 is 0.0176 mol.
Step 5: Convert the theoretical yield in moles to grams using the molar mass of Ag2SO4:
- The theoretical yield of Ag2SO4 is 0.0176 mol x 311.81 g/mol = 5.49 g
Step 6: Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100:
- Percent yield = (1.6 g ÷ 5.49 g) x 100 = 29.15%
Therefore, the percent yield of Ag2SO4 in this reaction is approximately 29.15%.
To calculate the percent yield, you need to compare the actual yield (1.6 g) to the theoretical yield, which is the amount of product that should have been obtained according to the balanced chemical equation.
First, determine the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To do this, you need to calculate the number of moles of each reactant.
The molar mass of AgNO3 (silver nitrate) is:
AgNO3 = 107.87 g/mol (Ag = 107.87 g/mol, N = 14.01 g/mol, O = 16.00 g/mol x 3)
The molar mass of Na2SO4 (sodium sulfate) is:
Na2SO4 = 142.04 g/mol (Na = 22.99 g/mol x 2, S = 32.06 g/mol, O = 16.00 g/mol x 4)
Using the given masses of AgNO3 and Na2SO4, calculate the number of moles of each reactant using the formula:
moles = mass / molar mass
For AgNO3:
moles of AgNO3 = 2.0 g / 169.87 g/mol
For Na2SO4:
moles of Na2SO4 = 2.5 g / 142.04 g/mol
Now, use the stoichiometry (mole ratio) from the balanced chemical equation to determine the theoretical yield of Ag2SO4 (silver sulfate) in grams.
From the balanced equation, you can see that the mole ratio between AgNO3 and Ag2SO4 is 2:1. This means that for every 2 moles of AgNO3, 1 mole of Ag2SO4 is produced.
Calculate the theoretical yield of Ag2SO4:
theoretical yield = (moles of AgNO3) x (molar mass of Ag2SO4)
Next, calculate the percent yield using the formula:
percent yield = (actual yield / theoretical yield) x 100
Now that you have the required formula and values, substitute them into the formula and calculate the percent yield.