4.500 g of aluminum sulfide and 4.500 g of water were reacted to form aluminum h

Question

4.500 g of aluminum sulfide and 4.500 g of water were reacted to form aluminum hydroxide and dihydrogen sulfide, as shown in the equation below. Al_2S_3 + 6 H_2O rightarrow 2 Al(OH)_3 + 3 H_2S What mass of Al(OH)_3 can be formed from the given amount of aluminum sulfide, if water was unlimited? Be sure to use proper significant figures and include a unit. What mass of Al(OH)_3 can be formed from the given amount of water, if aluminum sulfide was unlimited? Be sure to use proper significant figures and include a unit. Given 4.500 g of each reactant, what is the theoretical yield of aluminum sulfide for this reaction? Be sure to use proper significant figures and include a unit. If 3.564 g of aluminum sulfide was collected in the lab, what is the percent yield for this reaction?

Explanation / Answer

Ans. Balanced reaction: Al2S3 + 6 H2O -----> 2 Al(OH)3 + 3 H2S

Stoichiometry: 1 mol Al2S3 reacts with 6 mol H2O to form 2 mol Al(OH)3 and 3 mol H2S.

Moles of Al2S3 = Mass / molar mass = 4.500 g / (150.16 g mol-1) = 0.03 mol

Moles of H2O = Mass / molar mass = 4.500 g / (18.0 g mol-1) = 0.25 mol

#1. 1 mol aluminum sulfide (limiting reagent) forms 2 mol aluminum hydroxide.

            So, moles of Al(OH)3 produced = 2 x moles of Al2S3 in reaction mixture

                                                            = 2 x 0.03 mol = 0.06 mol

            Mass of Al(OH)3 produced = moles x molar mass

                                                = 0.06 mol x (78.00 g mol-1) = 4.68 g

#2. 6 mol H2O (limiting reagent) forms 2 mol aluminum hydroxide.

Moles of Al(OH)3 produced from 0.25 mol H2O = (1/6) x 0.25 mol = 0.04166 mol

            Mass of Al(OH)3 produced = moles x molar mass

                                                = 0.04166 mol x (78.00 g mol-1) = 3.25 g

#3. From #1,

            Moles of Al2S3 = 0.03 mol

            Moles of H2O = 0.25 mol

Molar ratio is Moles of Al2S3 : H2O = (0.03 : 0.25)

Multiply the molar ration with 24. Multiplying or diving a ration with nay factor does not affect the ratio. That is-

            Experimental molar ratio Al2S3 : H2O = (0.03 : 0.25) x 24 = 0.72 : 6

When compared with the normal stoichiometry ration of (Al2S3 : H2O = 1: 6) , the number of moles of Al2S3 is less than the stoichiometric value of 1.

So, Al2S3 is the limiting reactant.

Therefore, the moles of H2S (NOT aluminum sulfide as written in the question, make corrections at that point)

Now,

            Moles of H2S produced = 3 x moles of Al2S3 in reaction mixture                ; [see stoichiometry]

                                                            = 3 x 0.03 mol = 0.09 mol

Mass of H2S produced = mole x molar mass = 0.09 mol x (34.08 g mol-1) = 3.07 g

Theoretical yield, i.e. amount of H2S produced according to the given stoichiometry under theoretically optimum conditions (assuming 100% conversion of reactants into product) =3.07 g           

Now,

            % yield = (Mass of H2S obtained experimentally / Theoretical yield) x 100

                        = (3.564 g / 3.07 g) x 100 = 116.09 %

Note: % yield can’t exceed 100%. There might be some error with the provided values in the question.

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