Chapter 9

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HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 104 - <strong>Chapter</strong> 9 - (b) 2 NaOH + H 2 SO 4 ¡ Na 2 SO 4 + 2 H 2 O 10.0 g 10.0 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using H 2 O as the product: Since H 2 SO 4 produces less H 2 O, it is the limiting reactant and NaOH is in excess. 24. (a) 2 Bi(NO 3 ) 3 + 3 H 2 S ¡ Bi 2 S 3 + 6 HNO 3 50.0 g 6.00 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using Bi 2 S 3 as the product: (b) 110.0 g NaOH2a 1 mol 40.00 g b ¢ 2 mol H 2O 2 mol NaOH ≤ a 18.02 g b = 4.51 g H mol 2 O 110.0 g H 2 SO 4 2a 1 mol 98.09 g b ¢ 2 mol H 2O ≤ a 18.02 g b = 3.67 g H 1 mol H 2 SO 4 mol 2 O 150.0 g Bi(NO 3 ) 3 )a 1 mol 395.0 g ba 1 mol Bi 2S 3 ba 514.2 g b = 32.5 g Bi 2 mol Bi(NO 3 ) 3 mol 2 S 3 (6.00 g H 2 S)a 1 mol 34.09 g b ¢ 1 mol Bi 2S 3 3 mol H 2 S ≤ a 514.2 g b = 30.2 g Bi mol 2 S 3 Since H 2 S produces less Bi 2 S 3 , it is the limiting reactant and Bi(NO 3 ) 3 is in excess. 3 Fe + 4 H 2 O ¡ Fe 3 O 4 + 4 H 2 40.0 g 16.0 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using H 2 as the product: (40.0 g Fe)a 1 mol 55.85 g b ¢ 4 mol H 2 3 mol Fe ≤ a 2.016 g b = 1.93 g H mol 2 116.0 g H 2 O2a 1 mol 18.02 g b ¢ 4 mol H 2 4 mol H 2 O ≤ a 2.016 g b = 1.79 g H mol 2 Since H 2 O produces less H 2 , it is the limiting reactant and Fe is in excess. 25. Limiting reactant calculations C 3 H 8 + 5 O 2 ¡ 3 CO 2 + 4 H 2 O (a) Reaction between 20.0 g C 3 H 8 and 20.0 g O 2 Convert each amount to moles of CO 2 - 104 -
HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 105 - <strong>Chapter</strong> 9 - 120.0 g C 3 H 8 2a 1 mol 44.09 g b ¢ 3 mol CO 2 1 mol C 3 H 8 ≤ = 1.36 moles CO 2 120.0 g O 2 2a 1 mol 32.00 g b ¢ 3 mol CO 2 5 mol O 2 ≤ = 0.375 moles CO 2 O 2 is the limiting reactant. The yield is 0.375 moles CO 2 . (b) Reaction between 20.0 g C 3 H 8 and 80.0 g O 2 Convert each amount to moles of CO 2 120.0 g C 3 H 8 2a 1 mol 44.09 g b ¢ 3 mol CO 2 1 mol C 3 H 8 ≤ = 1.36 moles CO 2 180.0 g O 2 2a 1 mol 32.00 g b ¢ 3 mol CO 2 5 mol O 2 ≤ = 1.50 moles CO 2 C 3 H 8 is the limiting reactant. The yield is 1.36 moles CO 2 . (c) Reaction between 2.0 mol C 3 H 8 and 14.0 mol O 2 According to the equation, 2 mol C 3 H 8 will react with 10 mol O 2 . Therefore, C 3 H 8 is the limiting reactant and 4.0 mol O 2 will remain unreacted. 12.0 mol C 3 H 8 2¢ 3 mol CO 2 1 mol C 3 H 8 ≤ = 6.0 mol CO 2 produced (2.0 mol C 3 H 8 )¢ 4 mol H 2O 1 mol C 3 H 8 ≤ = 8.0 mol H 2 O produced When the reaction is completed, 6.0 mol CO 2 , 8.0 H 2 O, and 4.0 mol O 2 will be in the container. 26. The balanced equation is 2 C 3 H 6 + 9 O 2 ¡ 6 CO 2 + 6 H 2 O (a) Reaction between 15.0 g C 3 H 6 and 15.0 g O 2 . Convert each amount to moles of H 2 O 115.0 g C 3 H 6 2¢ 1 mol 42.08 g ≤¢6 mol H 2O 2 mol C 3 H 6 ≤ = 1.07 mol H 2 O 115.0 g O 2 2¢ 1 mol 32.00 g ≤¢6 mol H 2O 2 mol O 2 ≤ = 0.313 mol H 2O The O 2 is the limiting reactant. The yield is 0.313 mol H 2 O. - 105 -
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Chapter 9

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