Question

Can somebody please tell me if this is even correct?

Answer 1

typing...........

Answer 2

Thats quite alot of typing O.o

Answer 3

I found this question and answer off the internet because I wanted to see an example of calculating limited reagent and percent yield. I found this but I really don't understand it and I have no idea if it is even correct. If one of you could please rewrite it in a more understandable format and tell me if it is even correct or not, it would be a great help:)

question: A student used 7.15 g of CaCl2 and 9.25 g of K2CO3 to make CaCO3. The actual yield was 6.15 g of CaCO3. Calculate
the limiting reagent and the percent yield.


answer: 1 mole CaCI2 requires 1 mole K2CO3 to fully react A 1 : 1 reaction ratio moles CaCI2 = mass / molar mass = 7.15 g / 110.98 g/mol = 0.064426 moles moles K2CO3 = 9.25 g / 138.21 g/mol = 0.066927 mol The balanced equation shows that 1 mole CaCI2 reacts to form 1 mole CaCO3. Another 1 : 1 ratio So 0.064426 moles CaCI2 can form up to 0.064426 moles CaCO3 mass CaCO3 possible = moles x molar mass = 6.45 g ANSWER: % yield = actual mass / theoretical yield x 100/1 = 6.15 g / 6.45 g = 100/1 = 95.3% the reaction is 1:1, so the limiting reagent is going to be whatever you have less of, in this case the CaCI2.

Answer 4

math mate can u help me

Answer 5

The answer is valid, and in fact, reflects reality.
Since the molar ratio is 1:1, so the limiting reagent is whatever is less.... in terms of moles.
CaCl2: 7.15/111=0.064 mol
K2CO3: 9.25/138=0.067 mol
So the limiting reagent is CaCl2
The reason that there is only 6.14 g instead of 6.45 g could be due to experimental loss, e.g. some get stuck in the filter paper, some left in solution (<0.015g/L), etc. Since it is not a quantitative analysis, some loss is not critical.

Answer 6

Thanks @mathmate ! Now I understand it completely:D

Answer 7

You're welcome! :)
However, I suggest you post, in the future, chemistry questions in the chemistry section to give chemists a chance to help. You are always welcome to post a \(link\) here so that all discussions will be centralized there.