c101_f10_1027

c101_f10_1027 - Answers to practice problem: The recipe:...

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Answers to practice problem : The recipe: 2Al + 3Cl 2 Al 2 Cl 6 . The ingredients: 2.70 g Al and 4.05 g Cl 2 . What is the limiting reactant? Cl 2 (0.0571 moles Cl 2 will make 0.0190 moles Al 2 Cl 6 ) (or, 0.0571 moles Cl 2 requires 0.0381 moles Al, and we have 0.100 moles Al) (or, 0.100 moles Al requires 0.150 moles Cl 2 , and we don’t have enough Cl 2 ) How much product can be made? 0.0190 moles Al 2 Cl 6 (5.08 grams) How much excess reactant is leftover? 1.67 g Al (e.g., 5.08 + 1.67 = 6.75 = 2.70 + 4.05)
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Chemistry 101 Fall 2010 2 Combustion reaction Combination reaction with oxygen: C 12 H 22 O 11 (s) + 12 O 2 (g) 12CO 2 (g) + 11H 2 O(g) Carbon atoms combine with oxygen to form CO 2 molecules. Hydrogen atoms combine with oxygen to form H 2 O molecules. Phase notation (g) = gas (l) = pure liquid (s) = solid (aq) = aqueous = dissolved in water NaCl(s) Na + (aq) + Cl - (aq)
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3 For very soluble or reactive compounds, NaCl(aq) => Na + (aq) + Cl - (aq), and HCl(aq) => H + (aq) + Cl - (aq). For less reactive compounds,
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c101_f10_1027 - Answers to practice problem: The recipe:...

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