Whats the percent yield in the reaction:
P+5C2H5OH+ 5/2 I2 -> 5C2H5I + H3PO4 + H2O
if you have 4.14g phosporous, 27.5g ethanol, 63.5g iodine, and you get 73g ethyliodine (C2H5I)
This is another limiting reagent problem but this time you have THREE reactants. You will need to compare two of the reagent to find which is limiting, then compare that one with the third. The value you obtain at the end for grams is the theoretical yield.
%yld = (actual yield/theor yield)*100 =
DrBob222 can you please explain more specifically.I don't know if iodine or phosporus is the limiting reagent so i can compare it with C2H5I. Can you tell me only the process how to compare the 3 reagents to find the limiting reagent please.
To calculate the percent yield in a chemical reaction, you need to compare the actual yield (the amount of the desired product obtained experimentally) with the theoretical yield (the maximum amount of product predicted by stoichiometry calculations).
Here's the step-by-step procedure to determine the percent yield in this reaction:
1. Calculate the molar masses of the reactants and products:
- Phosphorus (P): 30.97 g/mol
- Ethanol (C2H5OH): 46.07 g/mol
- Iodine (I2): 253.80 g/mol
- Ethyliodine (C2H5I): 155.99 g/mol
2. Convert the given masses of reactants (phosphorus, ethanol, and iodine) to moles:
- Moles of phosphorus (P) = mass / molar mass = 4.14 g / 30.97 g/mol
- Moles of ethanol (C2H5OH) = mass / molar mass = 27.5 g / 46.07 g/mol
- Moles of iodine (I2) = mass / molar mass = 63.5 g / 253.80 g/mol
3. Identify the limiting reactant by comparing the stoichiometric ratios in the balanced equation:
- The balanced equation shows a mole ratio of 1 mole P to 5 moles C2H5OH to 5/2 moles I2.
- Calculate the moles of ethyliodine (C2H5I) that can be formed from each reactant using the stoichiometric ratios. The limiting reactant will be the one that produces the least amount of C2H5I.
4. Calculate the theoretical yield of ethyliodine (C2H5I) using the moles of the limiting reactant:
- Theoretical yield (moles of C2H5I) = Moles of limiting reactant * (5 moles C2H5I / 1 mole limiting reactant)
5. Convert the theoretical yield from moles to grams using the molar mass of ethyliodine (C2H5I):
- Theoretical yield (grams of C2H5I) = Theoretical yield (moles of C2H5I) * molar mass of C2H5I
6. Calculate the percent yield using the actual and theoretical yields:
- Percent yield = (Actual yield / Theoretical yield) * 100
Now, let's perform the calculations:
1. Molar masses:
- Phosphorus (P): 30.97 g/mol
- Ethanol (C2H5OH): 46.07 g/mol
- Iodine (I2): 253.80 g/mol
- Ethyliodine (C2H5I): 155.99 g/mol
2. Moles of reactants:
- Moles of phosphorus (P) = 4.14 g / 30.97 g/mol = 0.134 mol
- Moles of ethanol (C2H5OH) = 27.5 g / 46.07 g/mol = 0.597 mol
- Moles of iodine (I2) = 63.5 g / 253.80 g/mol = 0.250 mol
3. Limiting reactant:
- Based on the stoichiometry, the 5/2 moles of iodine (I2) require 5/2 * (5/2) = 6.25 moles of ethanol (C2H5OH), while only 0.250 moles of iodine (I2) are available.
- Therefore, iodine (I2) is the limiting reactant.
4. Theoretical yield of ethyliodine (C2H5I):
- Theoretical yield (moles of C2H5I) = 0.250 moles I2 * (5 moles C2H5I / 0.250 moles I2) = 5 moles C2H5I
5. Theoretical yield of ethyliodine (C2H5I) in grams:
- Theoretical yield (grams of C2H5I) = 5 moles C2H5I * 155.99 g/mol = 779.95 g
6. Percent yield:
- Percent yield = (Actual yield / Theoretical yield) * 100 = (73 g / 779.95 g) * 100 = 9.36%
Therefore, the percent yield in this reaction is 9.36%.