Which of the following is the best example of an object absorbing light? A black sweater in the sun

The normality that would be calculated will be to high because the change in volume will be greater than the actual change in volume. if the buret tip is not filled when reading the initial volume, the actual volume should be lesser with that reading. so if you will you the higher reading the change in volume or the volume you use in titration will be higher than the actual

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3 years ago

If 4.50 g of HCl are reacted with 15.00 g of Caco, according to the following balanced chemical equation, calculate the theoreti

Tom [10]

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:

$moles\ CO_{2}\ = (4.50\ g\ HCl)\(\frac{1\ mol\ HCl}{36.46094\ g})( \frac{1\ mol\ CO_{2} }{2\ mol\ HCl}) \\moles\ CO_{2}\ =\ 0.0617098$

For CaCO3:

$moles\ of\ CO_{2}\ = (15.00\ g\ CaCO_{3})\ (\frac{1\ mol\ CaCO_{3} }{100.0869\ g\ CaCO_{3} })\ (\frac{1\ mol\ CO_{2} }{1\ mol\ CaCO_{3} })\\moles\ of\ CO_{2}\ = \ 0.1499$

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.

$grams\ CO_{2}\ =\ (0.0617098\ mol\ CO_{2}) (\frac{44.01\ g\ CO_{2}}{1\ mol\ CO_{2}})\\grams\ CO_{2}\ =\ 2.71585$

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:

$percent\ yield\ of\ CO_{2}\ = (\frac{2.50\ g}{2.71585\ g}) (100)\\percent\ yield\ of\ CO_{2}\ = 92.05221$

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

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2 years ago