Chapter 11 Review
Define the following: Stoichiometry Mole ratio Excess reactant Limiting reactant Theoretical yield Actual yield Percent yield You can find all the definitions in the book. Look them up and STUDY!!!!
Find the following molar masses: O2 : 16 x 2 = 32 g/mol O : 15.999 g/mol AlPO4 : 26.982 + 30.974 + (15.999 x 3) = 105.95 g/mol NaCl : 22.990 + 35.453 = 58.443 g/mol C6H5Cl : (12.011 x 6) + (1.0079 x 5) + 35.453 = 112.5585 g/mol CuO : 63.546 + 15.999 = 79.539 g/mol
Determine the number of steps for the following problems (1, 2, or 3): Given the following equation: 2 KClO3 –> 2 KCl + 3 O2 How many moles of O2 can be produced by letting 12.00 moles of KClO3 react? 1 step How many grams of O2 can be produced by letting 34.7g of KClO3 react? 3 steps How many moles of O2 can be produced by letting 14.6g of KClO3 react? 2 steps
Create the following mole ratios: _4_Ag(s) + _2_H2S(g) + _1_O2(g) _2_Ag2S(s) + _2_H2O(l) (Equation must first be balanced.) Ag : H2S 4:2 = 2:1 O2 : Ag2S 1: 2 Ag2S : H2O 2:2 = 1:1 O2 : H2S 1:2 Ag : O2 4:1 H2O : H2S 2:2 = 1:1 How many ratios can this equation form?
What is the molar ratio of all elements/compounds shown? 2 : 3 : 2
Solve the following: Given the following equation: 2 KClO3 –> 2 KCl + 3 O2 How many moles of O2 can be produced by letting 12.00 moles of KClO3 react?
Solve the following: Given the following equation: 2 KClO3 –> 2 KCl + 3 O2 How many grams of O2 can be produced by letting 34.7g of KClO3 react?
Solve the following: Given the following equation: 2 KClO3 –> 2 KCl + 3 O2 How many moles of O2 can be produced by letting 14.6g of KClO3 react?
Solve the following: Given the following equation: Al2(SO3)3 + 6 NaOH 3 Na2SO3 + 2 Al(OH)3 If 10.0 g of Al2(SO3)3 is reacted with 10.0 g of NaOH to form Na2SO, determine the limiting reactant
Solve the following: Given the following equation: 2 FePO4 + 3 Na2SO4 1 Fe2(SO4)3 + 2 Na3PO4 What is the percent yield of this reaction if takes place with 25g of FePO4 and an excess of Na2SO4, and produces 18.5g of Fe2(SO4)3