F10HW5-CH4

# F10HW5-CH4 - EWORK SET 5 CHAPTER 4 3.10 The limiting...

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EWORK SET 5, CHAPTER 4 3.10 The limiting reactant is the reactant that is entirely consumed when the reaction is complete. Because the reaction stops when the limiting reactant is used up, the moles of product are always determined by the starting number of moles of the limiting reactant. 3.23 a. The limiting reactant would be the charcoal because the air would supply as much oxygen as needed. b. The limiting reactant would be the magnesium because the beaker would contain much more water than is needed for the reaction (approximately 18 mL of water is one mole). c. The limiting reactant would be the H 2 because the air could supply as much nitrogen as is needed. H 2 is much more expensive than N 2 . 3.91 {OCL} Focus: First determine whether KO 2 or H 2 O is the limiting reactant by calculating the moles of O 2 that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of O 2 formed. O H mol 2 O mol 3 x O H mol 0.15 2 2 2 = 0.22 5 mol O 2 2 2 2 KO mol 4 O mol 3 x KO mol 0.25 = 0.18 7 mol O 2 (KO 2 is the limiting reactant) The moles of O 2 produced = 0.19 mol. Alternatively , we can find the limiting reagent as follows: 4 KO 2 + 2 H 2 O 4 KOH + 3 O 2 Given moles 0.25 0.15 Ratio of moles present = 2 2 mol KO 0.25mol = mol H O 0.15mol = 1.67 Required ratio for reaction (see balanced equation): 2 2 mol KO 4 mol = mol H O 2 mol = 2.0 Ratio of moles present is too small , meaning that we do not have enough KO 2 . Hence KO 2 is the limiting reagent.

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3.94 a) First, determine whether CS 2 or O 2 is the limiting reactant by calculating the moles of SO 2 that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of SO 2 formed. Use the molar mass of SO 2 to calculate the mass of SO 2 formed. Then, calculate the mass of the unconsumed reactant. CS 2 + 3O 2 CO 2 + 2SO 2 30.0 g O 2 x 2 2 1 mol O 32.00 g O x 2 2 2 mol SO 3 mol O = 0.625 0 mol SO 2 35.0 g CS 2 x 2 2 1 mol CS 76.15 g CS x 2 2 2 mol SO 1 mol CS = 0.919 2 mol SO 2 O 2 is the limiting reactant. b) Mass SO 2 formed = 0.625 0 mol SO 2 x 2 2 SO mol 1 SO g 64.07 = 40.0 4 = 40.0 g c) CS 2 is left unconsumed at the end of the reaction. Firstly, calculate the mass of CS 2 that reacts with the given amount of the limiting reagent. Do this by converting from moles of O 2 to moles of CS 2 , then to mass of CS 2 : 30.0g O 2 x 2 2 2 2 2 1 mol O 1 mol CS 76.15g CS x x 32.0g 3 mol O 1 mol CS = 23.8g CS 2 Now we can find the mass of CS 2 which remains unreacted: The unreacted CS 2 = 35.0 g total CS 2 - 23.8 g reacted CS 2 = 11.2 g CS 2 4.2 A strong electrolyte is a soluble substance that exists in solution almost entirely as ions. An example is NaCl. When NaCl dissolves in water, it dissociates almost completely to
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## This note was uploaded on 02/20/2011 for the course CHEM 110 taught by Professor Dr.huston during the Spring '06 term at Pittsburgh.

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F10HW5-CH4 - EWORK SET 5 CHAPTER 4 3.10 The limiting...

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