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Quiz 4-26-12

Quiz 4-26-12. Answers. 2.30 g Na are reacted with 100.0 mL of 0.10 M phosphoric acid. Calculate the number of liters of hydrogen gas formed at STP. 6 Na + 2H 3 PO 4  3 H 2 + 2 Na 3 PO 4. 2.30 g Na. 1 mole Na. 3 mole H 2. 22.4 L H 2. = 1.12 L H 2. 23.0 g Na. 6 mole Na. 1 mole H 2.

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Quiz 4-26-12

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  1. Quiz 4-26-12 Answers

  2. 2.30 g Na are reacted with 100.0 mL of 0.10 M phosphoric acid. Calculate the number of liters of hydrogen gas formed at STP. 6 Na + 2H3PO4 3 H2 + 2 Na3PO4 2.30 g Na 1 mole Na 3 mole H2 22.4 L H2 = 1.12 L H2 23.0 g Na 6 mole Na 1 mole H2

  3. 2.30 g Na are reacted with 100.0 mL of 0.10 M phosphoric acid. Calculate the number of liters of hydrogen gas formed at STP. 6 Na + 2H3PO4 3 H2 + 2 Na3PO4 0.100 L H3PO4 0.1 mole H3PO4 3 mole H2 22.4 L H2 = 0.336 L H2 1L H3PO4 2 mole H3PO4 1 mole H2

  4. When you actually perform the experiment, you get 0.30 L of H2. What is the percent yield. • The two values for the volume of H2 were 0.336 L and 1.12 L of H2 • 0.336 L is the theoretical yield. 1.12 is just wishful thinking. • 0.30 L is the actual yield. Actual yield 0.30 L H2 X 100 % = 89% theoretical yield 0.336 L H2

  5. A 0.336 L of sample of hydrogen gas at STP is heated to 23oC at 745 mm Hg.Calculate the volume of the gas at room temperature. Identify the variables V1 = 0.336 L T1 = 273 K P1 = 760 mm Hg T2 = 23 + 273 = 296 K P2 = 745 mm Hg V2 = ? P1 V1 T2 = V2 P2 T1 (760 mm Hg) (0.336 L) (296 K) 0.372 L = (745 mm Hg) (273 K)

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