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UNIT 7 REVIEW

UNIT 7 REVIEW. CHEMISTRY 11 Apr. 16, 2010. Carla Manujyot Bikram Kevin Parveer Diane Allen Tanvir Kelsey Hardeep Jennifer Johnny Kauser Charan Jerry Jason Dawn Amanda Nhi Anna Mohammad Daman Munreet Nindy Manraj Rubal Sharon. Gurleen Camille Christabel

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UNIT 7 REVIEW

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  1. UNIT 7 REVIEW CHEMISTRY 11 Apr. 16, 2010

  2. Carla Manujyot Bikram Kevin Parveer Diane Allen Tanvir Kelsey Hardeep Jennifer Johnny Kauser Charan Jerry Jason Dawn Amanda Nhi Anna Mohammad Daman Munreet Nindy Manraj Rubal Sharon

  3. Gurleen Camille Christabel Kristine Dinesh Josh Dilraj Angelique Alson Belpreet Raman Baltej Puneet Manreet Uzair Ronald Nicole Harjit Aaron Sharan Tasha Nitin Kelly Navjeet Amanpreet

  4. Tashfiq Maria Monica Chakshoo Rajvinder Harman P. Thomas Julio Harjit Gagan Sandra Seema Avneet Eric Bobby Harman G. Onkar Sanam Atiq Preston Karen Louie Kanwar Alyssa Eric Hussain Ardaman Hammad

  5. Big Question 2C7H6O2 + 15O2 14CO2 + 6H2O Based on the chemical equation above, if you have 10.0 g C7H6O2 and 17.0 g O2, what mass of excess reactant will be left over when the reaction has finished? What mass of CO2 is formed?

  6. Big Answer 2C7H6O2 + 15O2 14CO2 + 6H2O (10.0 g C7H6O2)×(1 mol C7H6O2 / 122 g) =0.0820 mol C7H6O2 (0.0820 mol C7H6O2)×(15 mol O2 / 2 molC7H6O2) = 0.615 mol O2 (0.615 mol O2) × (32 g / 1mol O2) = 19.7 g O2 But we only have 17.0 g O2! Therefore, O2 is the limiting reactant (C7H6O2 is the excess reactant)

  7. Big Answer 2C7H6O2 + 15O2 14CO2 + 6H2O Stoichiometry follows the LIMITING reactant (17.0 g O2)×(1 mol O2 / 32 g) = 0.531 mol O2 (0.531 mol O2)×(2 mol C7H6O2 / 15 mol O2) =0.0708 mol C7H6O2 (0.0708 mol C7H6O2)×(122 g / 1 mol C7H6O2) =8.64 g 10.0 g C7H6O2 – 8.64 g C7H6O2 = 1.36 g C7H6O2

  8. Big Answer to Second Question 2C7H6O2 + 15O2 14CO2 + 6H2O (17.0 g O2)×(1 mol O2 / 32 g) = 0.531 mol O2 (0.531 mol O2)×(14 mol CO2 / 15 mol O2) =0.496 mol CO2 (0.496 mol CO2)×(44 g / 1 mol CO2) = 21.8 g CO2

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