Question

For each of the following reactions, calculate the grams of indicated product when 16.4 g of the first reactant and 10.1 g of the second reactant are used. You may want to reference (Pages 253 - 258) Section 7.9 while completing this problem.

Answer 1

1)

moles Al2S3 = 16.4 g / 150.1580772 g/mol = 0.10921823391 moles

moles H2O = 10.1 g / 18.01528 g/mol = 0.56063519412 moles

moles H2S produced = 0.56063519412 / 2 = 0.28031759706 moles

mass of H2S = 0.28031759706 mol x 34.08088 g/mol = 9.55347038736 g

Answer: Mass of H₂S = 9.55 g

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2) PART-D:

Are you sure Part-D is following up of above question, i mean relates to mass of H₂S??????

percent yield = (actual yield / theoretical yield) x 100%

percent yield = (186 g / 9.55 g) x 100% = 1947 %

Answer: percent yield = 1947 % ------------------> this is impossible in any reaction

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3) PART-B:

moles Ca(OH)2 = 330 g / 74.09268 g/mol = 4.4538812741 mol

energy absorbed = 4.4538812741 mol x 65.3 kJ = 291 kJ

energy absorbed =291 kJ

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AlzS3 (s) + 6H2O(1)——2Al(OH)3 (aq) + 3H2S(g) Calculate the mass of H2S. : PÅ P O a ? mass of (H2S) = 9.55 Submit Request Answer

Part D What is the percent yield if the actual yield from this reaction is 186 g ? Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s) C! Å 1947 O a? % This answer is on your own risk - please see description Submit Previous Answers

Part B Determine how many kilojoules are absorbed when 330 g of Ca(OH)2 reacts via the following reaction: Ca(OH)2 (s) + CaO(s) + H2O(1), AH = +65.3 kJ VO AQ R o ? 291 Submit Request Answer

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