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professor190 [17]
3 years ago
7

The student measures the actual pressure of the CO2(g) in the container at 425 K and observes that it is less than the pressure

predicted by the ideal gas law. Explain this observation.
Chemistry
1 answer:
Naddika [18.5K]3 years ago
4 0

Answer:

Answer in explanation

Explanation:

Gases are expected to behave as ideal gases at particular temperature and pressure know as the S.T.P.

At a temperature of 273k and a pressure of 1atm, it is expected that a mole of a gas will occupy a volume of 22.4L. Now, a particular student measured the pressure of a gas at 425K, it is expected that the pressure is different from the one that is supposed to be observed at STP and thus , the measures quite differently

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if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?
vlada-n [284]

Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al: 2 moles
  • H₂SO₄: 3 moles
  • Al₂(SO₄)₃. 1 mole
  • H₂: 3 moles

The molar mass of the compounds is:

  • Al: 27 g/mole
  • H₂SO₄: 98 g/mole
  • Al₂(SO₄)₃: 342 g/mole
  • H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Al: 2 moles ×27 g/mole= 54 grams
  • H₂SO₄: 3 moles ×98 g/mole= 294 grams
  • Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
  • H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield} x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know:

  • actual yield= 75.26 grams
  • theorical yield= 88.67 grams

Replacing in the definition of percent yields:

percent yield=\frac{75.26 grams}{88.67 grams} x100

Solving:

<u><em>percent yield= 84.88%</em></u>

Finally, the percent yield for the reaction is 84.88%.

Learn more about

the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

percent yield:

brainly.com/question/14408642

#SPJ1

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