siperaTeera
01-15-2020, 09:26 AM
calculate the maximum numbers of moles and grams of H2S that can form when 139 g of aluminum sulfide reacts with 116 g of water:Al2S3+ H2O -gt; Al (OH)3 + H2S (this is unbalanced)(a) mol H2S(b) g H2S© What mass of the excess reactant remains?_ g excess reactantPlease write your final answers with the correct amount of sig figs and please write out how you did it I need to know how to do this and I don\'t know which part I am messing up!You MUST start with a balanced equationThen, because the masses of two reactants are given, you need to determine which of the two is the limiting reagent i.ewhich you have the least amount ofThen finally use the quantity of the limiting reagent to find the quantity of H₂S that can be produced1.) Al₂S₃ + 6 H₂O → 2 Al(OH)₃ + 3 H₂S (molar ratio of Al₂S₃:H₂O is 1:6) 2.) determine limiting reagent: Al₂S₃ 139-g Al₂S₃ x ( 1 mol Al₂S₃/150.3-g Al₂S₃) 0.925 mol Al₂S₃ H₂O 116-g H₂O x ( 1molH₂O/18.0-g H₂O) 6.44 mol H20 The balanced equation shows 1 mole of Al₂S₃ requires 6 moles of H₂O Since we have more than 6 moles of water but less than 1 mole of aluminum sulfide, aluminum sulfide limits the product30.9248 mol Al₂S₃ x (3 mol H₂S / 1 mol Al₂S₃) x 34.12-g H₂S / 1 mol H₂S a.) 0.9248 mol Al₂S₃ x (3 mol H₂S / 1 mol Al₂S₃) 2.77 mol H₂S b.) 0.9248 mol Al₂S₃ x (3 mol H₂S / 1 mol Al₂S₃) x 34.12-g H₂S / 1 mol H₂S 94.7-g H₂S c.) since the water we needed was 6 times the moles of Al₂S₃, we used 6 x 0.9248 5.549 moles of water 5.549 mol H₂O x 18.0-g/mol 99.88-g usedThat leaves 116-g - 99.88 16.1-g remaining (sig digits allowed is 3 in each answer)Other related question