Chemistry 120 Name First Exam October 2, 2014 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1(10) 2(15) 3(18) 4(15) 5(10) 6(20) 7(6) 8(6) TOTAL 1) Please fill in the blanks, Atomic # # of Neut. # of Prot. # of Elect. Mass# 129 Te 2-52 2) Name the following compounds and circle whether they are soluble or insoluble. a) BaSO 4 Soluble Insoluble b) Fe 2 (CO 3 ) 3 Soluble Insoluble c) P 2 O 5 Soluble Insoluble d) (NH 4 ) 2 S Soluble Insoluble e) Zn(NO 3 ) 2 Soluble Insoluble
3a) Balance the following reactions i) Fe 3 S 4 + O 2 Fe 3 O 4 + SO 3 ii) Al + H 2 SO 4 Al 2 (SO 4 ) 3 + H 2 3b) Complete and balance the following reactions; iii) iv) CuCl 2 + Al 2 (SO 4 ) 3 Mg + HCl 3c) Predict the product and balance each of the following reactions; v) N 2 + S vi) Li + O 2 4) When a 9.474 g sample of an unknown organic acid is subjected to combustion analysis 13.895 grams of CO 2 and 5.684 g of H 2 O are produced. What is the empirical formula of the acid? 5) What is the concentration of H 3 PO 4 and if it takes 36.00 ml of 0.75 M NaOH to completely neutralize 25.00 ml of the H 3 PO 4?
6) When 60.00 ml of 0.6 M Fe(NO 3 ) 3 is mixed with 40.00 ml of 0.80 M K 2 S a black solid forms. How many grams of precipitate (solid) will form and what is the concentration of all the ions left in solution? What is the net ionic reaction? What is the concentration of all ions left in solution? How many grams of the black solid will be made? 7) Please propagate the error of the following mathematical operation and give the answer to the correct number of significant figures. (26.24 +/- 0.02) - ( 2.564 +/- 0.004) 35.00 +/- 0.02 8) Starting with solid CaCl 2 please describe how you would make 250 ml of 1.2 M CaCl 2.
Chemistry 120 Name Answer Key First Exam October 2, 2014 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit 1(10) 2(15) 3(18) 4(15) 5(10) 6(20) 7(6) 8(6) TOTAL 1) Please fill in the blanks, Atomic # # of Neut. # of Prot. # of Elect. Mass# 129 Te 2-52 77 52 54 129 52 2) Name the following compounds and circle whether they are soluble or insoluble. a) BaSO 4 barium sulfate Soluble Insoluble b) Fe 2 (CO 3 ) 3 iron (III) carbonate Soluble Insoluble c) P 2 O 5 diphosphorus pentoxide Soluble Insoluble d) (NH 4 ) 2 S ammonium sulfide Soluble Insoluble e) Zn(NO 3 ) 2 zinc nitrate Soluble Insoluble
3a) Balance the following reactions i) Fe 3 S 4 + 8 O 2 Fe 3 O 4 + 4 SO 3 ii) 2 Al + 3 H 2 SO 4 Al 2 (SO 4 ) 3 + 3 H 2 3b) Complete and balance the following reactions; iii) 3 CuCl 2 + Al 2 (SO 4 ) 3 3 CuSO 4 + 2 AlCl 3 iv) Mg + 2 HCl MgCl 2 + H 2 3c) Predict the product and balance each of the following reactions; v) N 2 + 5 S N 2 S 5 vi) 2 Li + O 2 2 Li 2 O 4) When a 9.474 g sample of an unknown organic acid is subjected to combustion analysis 13.895 grams of CO 2 and 5.684 g of H 2 O are produced. What is the empirical formula of the acid? 3 5 g CO 2 g l 3 5 l CO 2 3 5 l C 2 g l 3 g C 5 g H 2 O g l 3 5 l H 2 O 2 H H 2 O 3 l H g l 3 g H 9.474 g sample 3.7896 g C 0.6316 g H = 5.053 g oxygen (O) 5 53 g O g l 3 5 l O 3 5 l C 3 5 l O C O 3 l H 3 5 l O 2 H O e e e C H 2 O 5) What is the concentration of H 3 PO 4 and if it takes 36.00 ml of 0.75 M NaOH to completely neutralize 25.00 ml of the H 3 PO 4? M H + H + M OH OH ( 5 M N OH)( 3 L) (M H +) 25 L) M H + M H + u H 3 O 3H + H 3 O M H + 3 M H 3 O
6) When 60.00 ml of 0.6 M Fe(NO 3 ) 3 is mixed with 40.00 ml of 0.80 M K 2 S a black solid forms. How many grams of precipitate (solid) will form and what is the concentration of all the ions left in solution? What is the net ionic reaction? Overall: 2 Fe(NO 3 ) 3 + 3 K 2 S Fe 2 S 3 (s) + 6 KNO 3 Net Ionic: 2 Fe 3+ + 3 S 2- Fe 2 S 3 What is the concentration of all ions left in solution? Note: K + and NO 3 - are spectator ions Ion Moles of Fe(NO 3 ) 3 = (0.60 M)(0.060 L) = 0.036 mol Fe(NO 3 ) 3 Moles of K 2 S = (0.80 M)(0.040 L) = 0.032 mol K 2 S Moles Before Rxn Moles After Rxn Conc. Comments Fe 3+ 0.036 0.01467 0.01467 mol/0.100 L = 0.1467 M Fe 3+ - NO 3 0.108 0.108-0.108/0.100 L = 1.08 M NO 3 3 0.036 M = 0.108 M K + 0.064 0.064 0.064/0.100 L = 0.64 M K + There are more negative ions than positives. You need positives therefore Fe 3+ cannot be limiting. S 2- is limiting S 2-0.032 0 0/0.100 L = 0 M S 2- Limiting reactant Determine the limiting reactant Using charges: There are 0.108 mole of negative charges and 0.064 mol of positive charges so we have 0.44 mole too many negative charges. We need another 0.44 mole of positive charges and our source is Fe 3+ so there must be, + ge 3+ ge e Fe 3+ l Fe 3+ le ve S 2 i li i i g Standard method: compare the two reacting ions 2 Fe 3+ 3 S 2 l 32 l 2 33 le Fe 3+ ee e l Fe 3+ le ve S 2 i li i i g How many grams of the black solid will be made? Use the limiting reactant Fe 2 S 3 l 3 S 2 32 l le Fe 2 S 3 lefe 2 S 3 2 g l 2 2 g Fe 2 S 3
7) Please propagate the error of the following mathematical operation and give the answer to the correct number of significant figures. (26.24 +/- 0.02) - ( 2.564 +/- 0.004) 35.00 +/- 0.02 Propagate the subtraction: ( ) ( ) Therefore, 26.24 2.564 = 23.676 = 23.68 +/- 0.02 Propagate the division: ( ) ( ) So, e 2 e 2 e ig ig Answer = 0.6766 +/- 0.0007 8) Starting with solid CaCl 2 please describe how you would make 250 ml of 1.2 M CaCl 2. Note: CaCl 2 = 110.98 g/mol MV = moles = (1.2 M CaCl 2 ) (0.250 L) = 0.300 mole CaCl 2 0.300 mole CaCl 2 110.98 g/mol = 33.294 g CaCl 2 So, put some water into a 250 ml volumetric flask and then weigh out 33.294 g CaCl 2 and add it to the flask. Stir the mixture until all of the CaCl 2 dissolves and then add enough water to fill the flask to the 250 ml mark.