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Last Name 1 Student’s Name Professor’s Name Course Number Date CHEMISTRY QUESTION 24: 2NaHCO 3 => Na 2 CO 3 + H 2 CO 3 Molar mass of NaHCO 3 = 23 + 1 + 12.01 + (15.999 x 3) = 84.007 Molar mass of Na 2 CO 3 = (23 x 2) + 12.01 + (15.999 x 3) = 106.007 Molar mass of H 2 CO 3 = (1 x 2) + 12.01 + (15.999 x 3) = 62.007 a. Moles of NaHCO 3 = mass / Molar mass = 3.55 / 84.007 = 0.0423M For Na 2 CO 3 : For every 2 moles of NaHCO 3 we should theoretically obtain 1 mole of Na 2 CO 3 Theoretical yield of Na 2 CO 3 = (Molar mass of Na 2 CO 3 x Moles of NaHCO 3 ) / 2 = (106.007 x 0.0423) / 2 = 2.242g For H 2 CO 3 : For every 2 moles of NaHCO 3 we should theoretically obtain 1 mole of H 2 CO 3 Theoretical yield of H 2 CO 3 = (Molar mass of H 2 CO 3 x Moles of NaHCO 3 ) / 2 = (62.007 x 0.0423) / 2 = 1.311g

Last Name 2 Therefore, the theoretical yield of Na 2 CO 3 and H 2 CO 3 are 2.242g and 1.311g respectively. b. The product is Na 2 CO 3 . This is because, in the actual decomposition of NaHCO 3 , the measured mass of one of the products was 2.16g. Thus, the product is Na 2 CO 3 , as its theoretical yield is closer to the value measured in the actual decomposition. c. The percent yield= (actual yield/ theoretical yield) x 100 Actual yield= 2.16g Theoretical yield= 2.242g = (2.16g/ 2.242g) x 100 =96.343% The percent yield of Na 2 CO 3 is 96.343% QUESTION 25: In both systems, entropy generally increases. In System A, gases mixing leads to increased disorder, while in System B, the transition from a solid to both solid and gas increases overall molecular randomness, contributing to higher entropy.

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