Practice Problems 03 Key

# Practice Problems 03 Key - Disclaimer These Practice...

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Disclaimer: These Practice Problems provided to you do not in any way reflect the content of the actual exam. The main purpose of these problems is to acquaint you with previous lectures and to test your skills in the chapters covered so far. So, please do not rely solely on the Practice Problems if you really want to succeed in the exam. Good luck. PRACTICE PROBLEMS III ANSWER KEY 1) How many moles of aluminum hydroxide are formed when 4.35 moles of sodium hydroxide are reacted with an excess amount of aluminum sulfate? Al 2 (SO 4 ) 3 (aq) + 6 NaOH (aq) Î 2 Al(OH) 3 (s) + 3 Na 2 SO 4 (aq) 4.35 mol NaOH 2 mol Al(OH) 3 = 1.45 mol Al(OH) 3 6 mol NaOH 2) How many grams of Al 2 O 3 are produced by the reaction of 63.0 g Fe 2 O 3 ? Fe 2 O 3 (s) + 2 Al (s) Î Al 2 O 3 (s) + 2 Fe (s) 63.0 g Fe 2 O 3 1 mol Fe 2 O 3 1 mol Al 2 O 3 101.9 g Al 2 O 3 = 40.2 g Al 2 O 3 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 1 mol Al 2 O 3 3) According to a pre-laboratory theoretical yield calculation, a student’s experiment should have produced 1.44 g of magnesium oxide. However, he only obtained 1.23 g magnesium oxide when he weighed out his product. What is the student’s percent yield? % Yield: Actual yield X 100% Theoretical yield % Yield = 1.23 g = 85.4% 1.44 g x 100% 4) What is the theoretical mass of xenon tetrafluoride that should form when 131 g of xenon is reacted with 101 g of F 2 ? What is the percent yield if only 145 g of xenon tetrafluoride is actually obtained? Xe (g) + 2 F 2 (g) Î XeF 4 (s) 131 g Xe 1 mol Xe 2 mol F 2 38.00 g F 2 = 75.8 g F 2 131.3 g Xe 1 mol Xe 1 mol F 2 The calculated amount of F 2 is less than the given amount of F 2 (calculated < given), therefore F 2 is in excess amount and Xe is the limiting reactant. If you used F 2 instead of Xe: 101 g F 2 1 mol F 2 1 mol Xe 131.3 g Xe = 174 g Xe 38.00 g F 2 2 mol F 2 1 mol Xe

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The calculated amount of Xe is greater than the given amount of Xe (calculated > given), therefore Xe is the limiting reactant and F 2 is in excess amount. We can now calculate the theoretical yield of XeF
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## This note was uploaded on 04/06/2008 for the course BIO 1500 taught by Professor Pandolfi during the Spring '08 term at Wayne State University.

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Practice Problems 03 Key - Disclaimer These Practice...

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