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Quiz 4-27-2012

Quiz 4-27-2012 . Answers. 4.0 grams of calcium are reacted with 30.0 mL of 3.0 M acetic acid. How many liters of gas are formed (assume STP). 1 Ca + 2 HC 2 H 3 O 2  1 Ca(C 2 H 3 O 2 ) 2 + 1 H 2 40 g .03L 3.0 M ? L. 1 mole Ca 40 g Ca. 1 mole H 2 1 mole Ca.

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Quiz 4-27-2012

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  1. Quiz 4-27-2012 Answers

  2. 4.0 grams of calcium are reacted with 30.0 mL of 3.0 M acetic acid. How many liters of gas are formed (assume STP).1 Ca + 2 HC2H3O2 1 Ca(C2H3O2)2 + 1 H240 g .03L 3.0 M ? L 1 mole Ca 40 g Ca 1 mole H2 1 mole Ca 22.4 L H2 1 mole H2 4.0 g Ca = 2.2 L H2 3 moleHC2H3O2 1 L HC2H3O2 1 mole H2 2 mole HC2H3O2 22.4 L H2 1 mole H2 = 1.0 L H2 0.03 L HC2H3O2 1.0 L is actually formed, 2.2 L is just wishful thinking

  3. Find % yield if 0.90 L of H2 were actually formed. • Since 1.0 L of H2 were formed: actual 0.90 L H2 X 100 % X 100% = 90% yield theoretical 1.0 L H2

  4. Remember to change to Kelvins! A 3.0 L sample of gas at 100oC and 90 kPa is changed to STP. What is the new volume. Notice that the temperature, volume and pressure are changing. This tells you to use the combined gas law. P1V1 = P2V2 T1 T2 P1 V1 P2 V2 = T1 T2 Standard Temperature 273 K Standard Pressure 101.3 kPa 760 mm Hg 1 atm 90 kPa 3.0L 273 K =2.0 L P1 = 90 kPa T1 = 373 K V1 = 3.0 L P2 = 101.3 kPa T2 = 273 K V2 = ? 373K 101.3 kPa

  5. How many moles of gas are in a 3.0 L sample of a gas at 100oC and 90 kPa? Since the volume, temperature and pressure don’t change, we use PV = nRT n = ? V = 3.0 L T = 373 K P = 90 kPa R = ? P V = n RT (90 kPa)( 3.0 L) 8.31 L kPa / K mole (8.31 L kPa/K mole)( 373 K) R = 0.0821 Latm/ K mole R = 62.4 LmmHg / K mole R = 8.31 L kPa / K mole = 0.087 moles

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