what is the percentage yield for
Sr(NO3)2 + Na2SO4 -> SrSO4 + 2 NaNO3.
mass of strontium nitrate is 0.788g
mass of sodium sulfate is 1.258g
mass of dry filter paper 0.759 g
mass of dried precipitate in filter paper 1.317g
This is a two part question; the first part is to determine the amount of yield for 100%. This is a limiting reagent(LR) problem because amounts are given for BOTH reactants.
mols Sr(NO3)2 = grams/molar mass
mols Na2SO4 = grams/molar mass
Using the coefficients in the balanced equation, convert mols each to mols SrSO4. It is likely you will obtain two different values which means one is wrong. In LR problems the correct value is ALWAYS the smaller value and the reagent responsible for that number is the LR. Theh g SrSO4 = mols x molar mass and this is the theoretical yield(TY) (at 100%).
Then %yield = (actual yield/TY)*100 = ?
Note: Actual yield is 1.317-0.759 = ?
To calculate the percentage yield in a chemical reaction, you need to know the actual yield and the theoretical yield. The actual yield is the amount of product obtained from the reaction, while the theoretical yield is the amount of product that would be obtained in an ideal or perfect reaction.
1. Calculate the moles of Sr(NO3)2 and Na2SO4:
- Moles of Sr(NO3)2 = mass of Sr(NO3)2 / molar mass of Sr(NO3)2
- Moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4
2. Identify the limiting reactant:
- To determine the limiting reactant, compare the moles of each reactant to the stoichiometric ratio of the balanced equation. The reactant with the lower mole ratio is the limiting reactant, as it will be completely consumed in the reaction.
3. Calculate the theoretical yield of the desired product:
- Use the moles of the limiting reactant and the stoichiometry of the balanced equation to calculate the moles of the desired product.
- Theoretical Yield = Moles of limiting reactant * (Molar ratio of desired product / Molar ratio of limiting reactant)
4. Calculate the percentage yield:
- Percentage Yield = (Actual Yield / Theoretical Yield) * 100
Now let's perform the calculations:
1. Calculate moles of Sr(NO3)2:
- Molar mass of Sr(NO3)2 = (87.62 g/mol for Sr) + 2 * (14.01 g/mol for N) + 6 * (16.00 g/mol for O) = 211.63 g/mol
- Moles of Sr(NO3)2 = 0.788 g / 211.63 g/mol = 0.00372 mol
2. Calculate moles of Na2SO4:
- Molar mass of Na2SO4 = 2 * (22.99 g/mol for Na) + 32.06 g/mol for S + 4 * (16.00 g/mol for O) = 142.04 g/mol
- Moles of Na2SO4 = 1.258 g / 142.04 g/mol = 0.00886 mol
3. Determine the limiting reactant:
- According to the balanced equation, the mole ratio between Sr(NO3)2 and SrSO4 is 1:1, and between Na2SO4 and SrSO4 is 1:1.
- Since the mole ratio is the same for both reactants, we can compare the moles directly.
- Based on the calculations, we see that 0.00372 mol of Sr(NO3)2 is less than 0.00886 mol of Na2SO4. Therefore, Sr(NO3)2 is the limiting reactant.
4. Calculate the theoretical yield of SrSO4:
- The stoichiometric ratio between Sr(NO3)2 and SrSO4 is 1:1.
- Theoretical Yield = 0.00372 mol of Sr(NO3)2 * (1 mol of SrSO4 / 1 mol of Sr(NO3)2) = 0.00372 mol
5. Calculate the percentage yield:
- Actual Yield = mass of dried precipitate in filter paper - mass of dry filter paper = 1.317 g - 0.759 g = 0.558 g
- Percentage Yield = (0.558 g / 0.00372 mol) * 100 = 15012.90%
Therefore, the percentage yield of SrSO4 in the given reaction is approximately 15012.90%. It is quite unusual to have such a high percentage yield, so there might be some error in the measurements or calculations.