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UTA 1441 Final Answer Key 1. (43g Cu/1)(1 mole Cu/63.5 g)(1 mole CH 4 /4 moles Cu)(16g CH 4 /1 mole)= 2.7g 2. (1.5 mole C 2 H 4 /1)(3 moles O 2 /1 mole C 2 H 4 )(22.4 L O 2 /1 mole O 2 )= 100.8L 3. % = Mass Cl / Total Mass X 100 = 71g Cl / 85 gCH 2 Cl 2 X 100 = 83.5% 4. (48.2 g CH 4 /1)(1 mole CH 4 /16g CH 4 )(6.02 X 10 23 molecules CH 4 /1 mole) = 1.81 X 10 24 molecules CH 4 5. (65g C 2 H 2 O 2 /1)(32 g O / 58 g C 2 H 2 O 2 ) = 36 g Oxygen 6. (70.6 g C/1)(1 mole C / 12 g C)= 5.88 moleC/1.47 = 4 (5.9 g H/1)(1 mole H/1gH)= 5.90 mole H/1.47 = 4 (23.5 g O/1)(1 mole O/16g O)= 1.47/1.47 = 1 Empirical Formula is C 4 H 4 O. This has a molecular mass of 68amu. The true molecular mass of the compound is 136. 136/68 = 2, so the molecular formula for the compound is C 8 H 8 O 2 which has a molecular mass of 136 amu. 7. XeF 6 + 3 H 2 O XeO 3 + 6 HF (3g XeF 6 /1)(1 mole XeF 6 /245.3 g XeF 6 )(1 mole XeO 3 /1 mole XeF 6 ) (179.3g XeO 3 /1 mole XeO 3 ) = 2.19 g XeO 3 % yield = Actual / Theoretical X 100 = 1.2g / 2.2g X 100 = 54.5 % 8. 2H 2 + O 2 2 H 2 O (4g H 2 /1)(1 mole H 2 /2 g H 2 )(2 mole H 2 O/2 mole H 2 )(18g H 2 O/1 mole H 2 O) = 36g H 2 O (35gO 2 /1)(1 mole O 2 /32 g O 2 )(2 mole H 2 O/1 mole O 2 )(18g H 2 O/1 mole H 2 O) = 39.4 g The answer is 36g.

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