Question

Make UP lef Check my work Be sure to answer all parts. Calculate the maximum numbers of moles and grams of H2S that can form when 127.3 g of sulfide reacts with 143.0 g of water: Al2Sg + H2O ? Al(OH)3 + H2S [unbalanced] mol H2S g H2S What mass of the excess reactant remains? g excess reactant

Answer 1

Be sure to answer all parts. Calculate the maximum numbers of moles and grams of H,S that can f sulfide reacts with 143.0 g of water: form when 1273 g of aluminum Al Ss+ H20 AI(OE)3+ H2S [unbalanced] 2.54 mol H,s 86.7 gHS What mass of the excess reactant remains? 51.3 g excess reactant No. of moles of A12S3- (127.3 g) (150.14 g/mol0.848 mol No. of moles of H20 (143.0 g)/(18.02 g/mol) 7.93 mol Balanced chemical equation Al2S3 +6 H20 No. of moles of H20 needed to react with 0.848 mol of A12S3- (6/ 1) x 0.848- 5.08 mol Here; limiting reagent is A12S3 (since present in limited quantity) No. of moles of H2S formed = (3/ 1) x 0.848 = 2.54 mol weight of H2S formed (2.54 mol) x (34.08 g/mol)-86.7 grams Excess reactant (water) remain -(7.93 mol) (5.08 mol 2.85 mol weight of excess water (2.85 mol) x (18.02 g/mol) 51.3 g 2 AI(OH)3+ 3 H2S

Eater your answer in the provided box. Two successive reactions, A B and BC, have yields of 63% and 39%, respectively, what is the overall percent yield for conversion of A to C? 24.57 % yield overall yield-100x (63 / 100) x (39 / 100)-24.57 %