When 14.5g of so2 react with 21g of O2 what will be the theoretical and percentage yiad of the reaction if the actealyid of the reaction is 12g
The balanced chemical equation for the reaction between SO2 and O2 is:
2 SO2 + O2 -> 2 SO3
From the equation, we can see that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
First, we need to calculate the limiting reagent to determine the theoretical yield.
The moles of SO2 present = 14.5 g / 64.06 g/mol = 0.226 moles
The moles of O2 present = 21 g / 32 g/mol = 0.656 moles
From the above values, we see that O2 is in excess as it has more moles than required for the reaction. Therefore, SO2 is the limiting reagent.
Theoretical yield = (0.226 mol SO2) x (2 mol SO3/2 mol SO2) x (80.06 g/mol SO3) = 18.1 g SO3
Percentage yield = (actual yield / theoretical yield) x 100
Percentage yield = (12 g / 18.1 g) x 100 = 66.3% (approx)
To find the theoretical yield and percentage yield of a reaction, we first need to determine the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between SO2 and O2 is:
2 SO2 + O2 -> 2 SO3
According to the equation, 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
Step 1: Calculate the number of moles of SO2 and O2:
Moles of SO2 = mass / molar mass
Molar mass of SO2 = 32.07 g/mol
Moles of SO2 = 14.5 g / 32.07 g/mol = 0.452 mol
Moles of O2 = mass / molar mass
Molar mass of O2 = 32.00 g/mol
Moles of O2 = 21 g / 32.00 g/mol = 0.656 mol
Step 2: Determine the limiting reactant:
The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product formed. To determine the limiting reactant, we compare the moles of each reactant to the stoichiometry of the balanced equation.
According to the stoichiometry of the equation, 2 moles of SO2 react with 1 mole of O2. Thus, the ratio of SO2 to O2 is 2:1.
Moles of SO2 / Ratio = 0.452 mol / 2 = 0.226 mol
Moles of O2 / Ratio = 0.656 mol / 1 = 0.656 mol
From the calculations, we can see that the moles of SO2 (0.226 mol) are less than the moles of O2 (0.656 mol), meaning SO2 is the limiting reactant.
Step 3: Calculate the theoretical yield:
According to the balanced equation, 2 moles of SO2 produce 2 moles of SO3.
Moles of SO2 = Moles of SO3
0.226 mol of SO2 -> x mol of SO3
Since the stoichiometry is 1:1 for SO2 to SO3, we can conclude that the theoretical yield of SO3 is also 0.226 mol.
The molar mass of SO3 is 80.06 g/mol. Therefore, the theoretical yield of SO3 in grams is:
Theoretical yield = number of moles × molar mass
Theoretical yield = 0.226 mol × 80.06 g/mol = 18.09 g
Step 4: Calculate the percentage yield:
Percentage yield = (actual yield / theoretical yield) × 100
Given that the actual yield is 12 g:
Percentage yield = (12 g / 18.09 g) × 100 = 66.29%
Therefore, the theoretical yield of the reaction is 18.09 g, and the percentage yield is approximately 66.29%.