1 (103)

Prévia do material em texto

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS 
 
102 
 
 2
2
2 mol HClMoles of HCl reacted 0.0137 mol Mg(OH) 0.0274 mol HCl
1 mol Mg(OH)
= × = 
 
 mol solute 0.0274 mol HCl
0.035 mol/L
= = =Volume of HCl 0.78 L
M
 
 
4.112 The balanced equations for the two reactions are: 
 
 X(s) + H2SO4(aq) ⎯⎯→ XSO4(aq) + H2(g) 
 
 H2SO4(aq) + 2NaOH(aq) ⎯⎯→ Na2SO4(aq) + 2H2O(l) 
 
 First, let’s find the number of moles of excess acid from the reaction with NaOH. 
 
 32 4
2 4
1 mol H SO0.500 mol NaOH0.0334 L 8.35 10 mol H SO
1 L soln 2 mol NaOH
−× × = × 
 
 The original number of moles of acid was: 
 
 2 4
2 4
0.500 mol H SO
0.100 L 0.0500 mol H SO
1 L soln
× = 
 
 The amount of sulfuric acid that reacted with the metal, X, is 
 
 (0.0500 mol H2SO4) − (8.35 × 10−3 mol H2SO4) = 0.04165 mol H2SO4. 
 
 Since the mole ratio from the balanced equation is 1 mole X : 1 mole H2SO4, then the amount of X that 
reacted is 0.04165 mol X. 
 
 1.00 g X
0.04165 mol X
= =molar mass X 24.0 g/mol 
 
 The element is magnesium. 
 
4.113 Add a known quantity of compound in a given quantity of water. Filter and recover the undissolved 
compound, then dry and weigh it. The difference in mass between the original quantity and the recovered 
quantity is the amount that dissolved in the water. 
 
4.114 First, calculate the number of moles of glucose present. 
 
 0.513 mol glucose 60.0 mL 0.03078 mol glucose
1000 mL soln
× = 
 
 2.33 mol glucose 120.0 mL 0.2796 mol glucose
1000 mL soln
× = 
 
 Add the moles of glucose, then divide by the total volume of the combined solutions to calculate the molarity. 
 
 60.0 mL + 120.0 mL = 180.0 mL = 0.180 L 
 
 (0.03078 0.2796) mol glucose 1.72 mol/L
0.180 L
+
= = =Molarity of final solution 1.72 M 
 
4.115 First, you would accurately measure the electrical conductance of pure water. The conductance of a solution 
of the slightly soluble ionic compound X should be greater than that of pure water. The increased 
conductance would indicate that some of the compound X had dissolved.