Question

Work each of the following problems. SHOW ALL WORK TO RECEIVE ANY CREDIT (Hint: Make sure your chemical equations are balanced.) 1. When 18.0 g of O2 reacts with 25.0 g of CHSOH, what is the limiting reactant? What is the theoretical yield in grams of CO2? C2HsOH +O2 CO2 +H2On 2. What is the limiting reactant when 126 g of Ag reacts with 142 g of HNO,? What is the theoretical yield in grams of AgNO3? AgNO (a) +NO+H2O HNO (aq) Ag 3. When 2.4 moles of Mg is ignited in 1.5 moles of pure oxygen, what is the limiting reactant? What is the theoretical yield of MgO? Mge) O2( MgO()

Answer 1

1) C2H5OH + 3O2 -> 2CO2 + 3H2O

According to the balanced reaction,

1 mole C2H5OH require 3 mole O2

46 g C2H5OH require 96 g O2

25 g C2H5OH require 96×25/46 = 52.17 g O2

Since we have less than this O2 thus O2 is limiting reagent.

According to the reaction,

96 g O2 give 88 g CO2

18 g O2 give 88×18/96 = 16.5 g CO2

2) 3Ag + 4HNO3 -> 3AgNO3 + NO + 2H2O

4 mole HNO3 require 3 mole Ag

4×63 = 252 g HNO3 require 3×107.86 = 323.58 g Ag

142 g HNO3 require 323.58×142/252 = 182.33 g Ag

Since we have less than this Ag thus, Ag is limiting reagent.

According to the reaction,

323.58 g Ag give 509.61 g AgNO3

126 g Ag give 509.61×126/323.58 = 198.43 g AgNO3

3) Mg + 1/2 O2 -> MgO

According to the reaction,

1/2 mole O2 require 1 mole Mg

1.5 mole O2 require 2×1.5 = 3 mole Mg

Since we have less Mg thus, Mg is limiting reagent

1 mole Mg give 1 mole MgO

1 mole Mg give 40 g MgO

2.4 mole Mg give 40×2.4 = 96 g MgO

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