Question

aluminum sulfide and water react to form aluminum hydroxide and hydrogen sulfide gas. suppose 128.00 g of aluminum are added to 154.45 mL of water(d=1.00 g/ml). Answer the following questions.

a).before the reaction, how many water molecules are present in the reaction mixture?

b) when the reaction is finished. what mass, in g, of aluminum hydroxide can theoretically be produced?

c) limiting reactant?

d) mass in grams of remaining excessreactant if the theoretical yield of aluminum hydroxide is obtained?

e)if the percent yield is 87.6% under these conditions what is the actual yiel of product aluminum hydroxide?

please show how you solve

Answer 1

Al2S3 + 6H2O ---> 2Al(OH)3 + 3H2S is balanced equation

a) H2O mass = volume x density = 154.45 x 1= 154.45 g

H2O moles = mass of H2O / molarmass of H2O = 154.45/18.01528 = 8.57328

Number of H2O molecules = moles of H2O x avagdronumber

     = 8.57328 x 6.023 x10^23 = 5.1637 x 10^24

b) Al2S3 moles = mass of Al2S3 / molar mass of Al2S3 = 128/150.158 = 0.852435

as per reaction 1 Al2S3 reacts with 6H2O hence for 0.852435 moles Al2S3 moles of H2O needed = 6 x 0.852435 = 5.1146 but we had 8.573 moles H2O , hence H2O is excess reagent

c) Al2S3 is limiting reagent,

Al(OH)3 moles = 2 x Al2S3 moles = 2 x 0.852435 = 1.70487

b) Al(OH)3 mass = moles of AL(OH)3 x molar massof Al(OH)3 = 1.70487 x 78 = 132.98 g is theoretically calculated or expected mass

d) H2O moles reacted = 5.1146 , H2O moles left = ( 8.57328 - 5.1146) = 3.4587

H2O mass = moles of H2O x molar mass of H2O = 3.4587 x 18.01528 =62.3 g water is left

e) percent yiled = 100 x ( actual yield / theoretically yiled)

87.6 = 100 x ( actual yiled / 132.98)

actual yiled = 116.5 g