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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
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter3: Molecules, Moles, And Chemical Equations
Section: Chapter Questions
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.
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
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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