Which statement best describes the effects chemicals may have on the
2 answers:
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Answer: HCl is the limiting reactant and the theoretical yield is 2.72 g of CO2. If the actual yield was 2.50 g then, the percent yield is 92.0% when rounding off is done only for the final answer.
Further Explanation:
In order to determine the theoretical yield and the percent yield of CO2, the following steps must be done:
- Determine the limiting reactant. This is the reactant that will determine the amount of CO2 that will actually form.
- Determine the theoretical yield for CO2 when the limiting reactant is used.
- Get the percent yield by getting the ratio of the actual yield stated in the problem and the calculated theoretical yield multiplied by 100.
Determining the Limiting Reactant
The Limiting Reactant (LR) will produce fewer moles of the products. To check which of the reactants HCl or CaCO3 is the LR, we do dimensional analysis:
For HCl:
For CaCO3:
Since HCl produces fewer moles of CO2, then it is the limiting reactant. We will use the given amount to determine the theoretical yield for CO2.
Determining the Theoretical Yield
From Step 1, we know that 0.0617098 moles of CO2 will be produced. We will just convert this to grams.
Since the answer only requires 3 significant figures, the final answer is 2.72 grams CO2.
Determining the Percent Yield
Dividing the actual yield by the theoretical yield will give us the percent yield, which is an indicator of how efficient the experiment or the method used was.
From the problem, the actual yield was 2.50 g, hence, the percent yield is:
Rounding off to three significant figures, the percent yield is 92.0%. This suggests that the method used is somewhat efficient in producing CO2.
Learn More
- Learn More about Limiting Reactant brainly.com/question/7144022
- Learn More about Excess Reactant brainly.com/question/6091457
- Learn More about Stoichiometry brainly.com/question/9743981
Keywords: stoichiometry, theoretical yield, actual yield
Hello!
The dissociation reaction for Benzoic Acid is the following:
C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺
The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:
![Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X} (assume: 0,40-X\approx0,40)\\ \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\ \\ pH=-log([H_3O^{+}]=2,29](https://tex.z-dn.net/?f=Ka%3D%20%5Cfrac%7B%5BC_6H_5COO%5E%7B-%7D%5D%2A%5BH_3O%5E%7B%2B%7D%5D%20%7D%7B%5BC_6H_5COOH%5D%7D%3D%5Cfrac%7BX%2AX%20%7D%7B0%2C40%20-X%7D%20%20%28assume%3A%200%2C40-X%5Capprox0%2C40%29%5C%5C%20%20%5C%5C%20X%3D%20%5Csqrt%7B0%2C40%2A6%2C50%2A10%5E%7B-5%7D%20%7D%3D0.00510M%20%5C%5C%20%20%5C%5C%20pH%3D-log%28%5BH_3O%5E%7B%2B%7D%5D%3D2%2C29%20%20%20)
So, the pH of this Benzoic Acid solution is 2,29
Have a nice day!
The dissociation reaction for Benzoic Acid is the following:
C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺
The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:
So, the pH of this Benzoic Acid solution is 2,29
Have a nice day!