Liquid hexane (CH, 3(CH,) CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO,) and gaseous water (H,0). Suppose 6.03 g 4 of hexane is mixed with 31.g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Liquid hexane (CH, 3(CH,) CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO,) and gaseous water (H,0). Suppose 6.03 g 4 of hexane is mixed with 31.g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Chemistry for Engineering Students
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Chapter3: Molecules, Moles, And Chemical Equations
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Problem 3.115PAE: 3.115 The average person exhales 1.0 kg of carbon dioxide in a day. Describe how you would estimate...
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![Liquid hexane (CH3(CH,)¸CH;)
will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO,) and gaseous water (H,O). Suppose 6.03 g
of hexane is mixed with 31. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to 2
significant digits.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb5c430b1-26cf-45f4-9b69-3655cbe0cdf0%2F278be8cb-92db-40a5-8883-18a8b76f8b48%2Frcvqqtu_processed.png&w=3840&q=75)
Transcribed Image Text:Liquid hexane (CH3(CH,)¸CH;)
will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO,) and gaseous water (H,O). Suppose 6.03 g
of hexane is mixed with 31. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to 2
significant digits.
Expert Solution
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Step 1
The reaction taking place can be written as (C6H14 = CH3(CH2)4CH3 = hexane),
=> C6H14 + O2 -------> CO2 + H2O
Given : Mass of C6H14 taken = 6.03 g
And mass of O2 taken = 31 g
Molar mass of O2 = 32 g/mol
And molar mass of C6H14 = Atomic mass of C X 6 + Atomic mass of H X 14 = 86 g/mol
Since mass = moles X molar mass
=> 6.03 = moles of C6H14 X 86
=> moles of C6H14 taken = 0.0701 mol approx.
And 31 = moles of O2 taken X 32
=> moles of O2 taken = 0.96875 mol.
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