Answer:
Explanation:
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In this case, since the decomposition of sodium hydrogen carbonate is:
Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:
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Answer:
1) 17.5 mL
Explanation:
Hello,
In this case, the reaction between sulfuric acid and potassium hydroxide is:
In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:
And in terms of concentrations and volumes:
Thus, we solve for the volume of acid:
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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:
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.
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Keywords: stoichiometry, theoretical yield, actual yield