after balancing the equation,
Al2S3 + 6H2O 2 Al(OH)3 + 3 H2S
So, 1 mole of Al2S3 will react with 6 moles of H2O
i.e., 150.16g of Al2S3 will react with 108g(18×6) of H2O
So, 158g of Aluminium sulfide will react with=(108÷150.16)×158 = 113.6g of H2O
Here the limiting reagent is Aluminium sulfide.
So the excess reagent i.e., H2O = Given mass of excess reagent - required mass of that reagent = 131 - 113.6 = 17.4g of H2O
1mole ie, 150.16g of aluminium sulfide will give 3 moles of H2S
158 ie, (158/150.16 = 1.05 moles) of Al2S3 will give= 3×1.05= 3.15 moles of H2S
1 mole os H2S = 34.1g
3.15 molesof H2S = no of moles × molar mass = 3.15 × 34.1 =107.41g of H2S
Worksheet for quiz 2 (chapters 2 and 3) 1. (3 points) Calculate the maximum number of...
Calculate the maximum numbers of moles and grams of H2S that can form when 153.0 g of aluminum sulfide reacts with 143.0 g of water: Al2S3 + H2O → Al(OH)3 + H2S [unbalanced] ______ mol H2S ______ g H2S What mass of the excess reactant remains? ______ g excess reactant
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
2 attempts left Check my work Be sure to answer all parts. Calculate the maximum numbers of moles and grams of HS that can form when 167.3 g of aluminum sulfide reacts with 143.0 g of water: Al,S3 + H,0 - Al(OH)3 + H2S (unbalanced) 3.34 14,5 133.9 XL, What mass of the excess reactant remains? 41.31 Xxcess reactant
Calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 679g of iodine trichloride reacts with 115.6 g of water: _________ mol _________ g ICI3 + H2O ---> ICl + HIO3+ HCl [unbalanced] What mass of the excess reactant remains? What is that reactant? __ ICl3 __ H2O
Calculate the maximum numbers of moles and grams of iodic acid (HlO3) that can form when 338 g of iodine trichloride reacts with 159.7 g of water: ICl3 + H2O → ICl + HIO3 + HCl (unbalanced) _____ mol HlO3 _____ g HlO3 What mass of the excess reactant remains? _____ g
Calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 547 g of iodine trichloride reacts with 224.5 g of water: ICl3 + H2O → ICl + HIO3 + HCl [unbalanced] ___mol HIO3 ___g HIO3 What mass of the excess reactant remains?____g
please help with all questions! Fall 2109 CHE 151-01 Exam 2 Chapter 4-6 Question 5 Question 6 (21 Points) Al2S3 + H2O - Al(OH)3 + H2S Suppose 316.0 g aluminum sulfide reacts with 0.493 kg of water. a) Write the balanced equation (2 Points) b) What % of Al(OH)3 is Al by mass? (2 Points) c) Identify the limiting and excess reagents (2Points) d) Calculate the number of moles and the mass in (g) after the reaction is complete for:...
hi! can you please help me figure out part C and D please? I re-did my work on part A, but don't know how to do part C and D. Thank you so much :) H23 - 1.00794 (2) + 32.066 = 34.08188 Alz S3 > 26.981538 (2) + 32.066(3) = 150.161076 PART II: PROBLEMS: Give a complete solution to each problem. SHOW YOUR WORK! Follow proper procedures concerning units and sig figs. Point values given. 6. Consider this unbalanced...
Calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 431 g of iodine trichloride reacts with 214.2 g of water: ICly+ Ho ICI+HIO+HCI (unbalanced mol HIO3 HIO What mass of the excess reactant remains?
184.55 g of aluminum sulfide (50.2 g/mol) and 134.05 g of water (18.0 g/mol) react until all the limiting reactant is used. What is the mass of H2S (34.1 g/mol) that can be produced from these reactants? 1 Al2S3 + 3 H2O -----> 2 Al(OH)3 + 3 H2S