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Mrac [35]
3 years ago
14

If the measured gas pressure in a collection flask during a gas-forming reaction was determined to be 1.44 atm? If the measured

gas pressure in a collection flask during a gas-forming reaction was determined to be 1.44 atm, and the atmospheric pressure that day was measured as 0.95 atm, what is the partial pressure of the gas produced by the reaction?
Chemistry
1 answer:
ryzh [129]3 years ago
6 0
With the information given most likely in order to find the partial pressure of the gas produced you have to subtract the total air pressure in the collection flask by the atmospheric pressure since you assume that the flask started with atmospheric pressure when it was sealed and then the gas was added as the reaction took place increasing the pressure.
1.44atm-0.95 atm=0.49atm
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What is the Percent yield is 4.65 got copper is produced by 1.87 percent yield is 4.65 got copper is produced when 1.87 grams of
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Answer:

Percent yield = 69.4%

Explanation:

Given data:

Percent yield  = ?

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Solution:

Chemical equation:

2Al + 3CuSO₄   →   Al₂(SO₄)₃ + 3Cu

Number of moles of Aluminium:

Number of moles = mass/ molar mass

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Number of moles = 0.07 mol

Now we will compare the moles of Al with copper.

                 Al           :            Cu

                   2           :              3

                0.07         :              3/2×0.07 =0.105 mol                    

Mass of Copper:(Theoretical yield)

Mass = Number of moles × molar mass

Mass = 0.105 mol × 63.5 g/mol

Mass = 6.7 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 4.65 g/ 6.7 g × 100

Percent yield = 69.4%

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