Answer:
Percentage Yield is given as,
%age Yield = Actual Yield / Theoretical Yield × 100
This shows that the %age yield is directly depending upon the actual yield. And most of the time the percentage yield is less than 100 % because of the following factors.
Impure Starting Materials:
If the starting materials (reactants) are not pure then reaction will not completely form the desired product. Different by products will form which will decrease the %age yield.
Incomplete Reactions:
Not all reactions go to completion. In many reactions the starting material after some time stops forming the product due to different conditions. Some reactions attain equilibrium and stop increasing the amount of product. While, in some reactions a by products (like water) formed often react with the product to give a reverse reactions. Hence, the chemistry of reactions also causes the decrease in %age yield.
Handling:
Another major reason for decrease in yield is handling the product. Always some of the product is lost during the workup of the reaction like, taking TLC, doing solvent extraction, doing column chromatography, taking characterization spectrums. So, we can conclude that the %age yield will always be less than 100%.
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
The amount of matter in an object is its Mass...
Answer:
A
Explanation:
the Molar mass will be smaller as the content of the container is not directly proportional to the temperature of the water bath.
Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
