s 1, 2 and 3. Each question has 10 blanks. Please fill those blanks with the letters of the correct answers. Choices are given below each question. Thanks!

Question 1 (3 points) 100.0 mL of HCI required 25.00 mL of 0.2000 M NaOH to reach the end point. What is the concentration of the HCI ? The neutralization equation is HCI + NaOH - NaCl + H20 There is enough information to calculate the moles of NaOH 1__ mol NaOH (2__ L NaOH)(--- -) - _3_ mol NaOH 1 L Then use stoichiometry to convert moles of NaOH into moles of HCI 4__ mol HCI (.5__ mol NaOH)- - ---) " _6__ mol HCI _7__ mol NaOH We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI 8__ mol HCI concentration - -_9__ mol/L _10__ L HCI A. acid B. base C. HCI D. NaOH E. NaCI F. H20 G. 25.00 H. 0.02500 1. 0.2000 J. 1.000 K. 0.005000 L. 100.0 M. 0.1000 N. 0.05000 O. 2.000 P. 0.01000 Q. 0.001000 R. 58.31 S. 5.831 Question 2 (3 points) 100.0 mL of HCI required 25.00 mL of 0.2000 M Mg(OH)2 to reach pH 7. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)2 - MgCl2 + 2 H20 There is enough information to calculate the moles of Mg(OH)2 1_ mol Mg(OH)2 (2__ L Mg(OH)2 (- --) = __3__ mol Mg(OH)2 1 L Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI 14_ mol HCI ( 5__ mol Mg(OH)2)- --) - __6__ mol HCI __7__ mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI 8__ mol HCI concentration = _9__ mol/L _10__ L HCI A. acid B. base C. HCI D. NaOH E. NaCI F. H20 G. 25.00 H. 0.02500 1. 0.2000 J. 1.000 K. 0.005000 L. 100.0 M. 0.1000 N. 0.05000 O. 2.000 P. 0.01000 Q. 0.001000 R. 58.31 S. 5.831 Question 3 (4 points) 100.0 ml of HCI required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)2 - MgCl2 + 2 H20 There is enough information to calculate the moles of Mg(OH)2 _1_ _ mol Mg(OH)2 (5.831 g Mg(OH)2 (--- ---)= _2__ mol Mg(OH)2 __3__ g Mg(OH)2 Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI 4__ mol HCI (5__ mol Mg(OH)2)(--- ---) = __6_ mol HCI _7__ mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI 8_mol HCI concentration - 19__ mol/L 10_LHCI A. acid B. base C. HCI D. NaOH E. NaCI F. H2O G. 25.00 H. 0.02500 1. 0.2000 J. 1.000 K. 0.005000 L. 100.0 M. 0.1000 N. 0.05000 O. 2.000 P. 0.01000 Q. 0.001000