1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g)
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1 1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) A) 1 B) 2 C) 4 D) 5 E) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) balance O Al (s) + H 2 O (l)? Al(OH) (s) + /2 H 2 (g) balance O and H 2 Al (s) + 6 H 2 O (l)? 2 Al(OH) (s) + H 2 (g) eliminate fraction 2. When the following equation is balanced, the coefficient of O 2 is. C 2 H 4 O + O 2? CO 2 + H 2 O A) 2 B) 1 C) 5 D) E) 4 C 2 H 4 O + O 2? 2 CO 2 + H 2 O balance C C 2 H 4 O + O 2? 2 CO H 2 O balance C & H C 2 H 4 O + 5/2 O 2? 2 CO H 2 O balance C, H & O 2 C 2 H 4 O + 5 O 2? 4 CO H 2 O eliminate fraction. Of the following compounds, which one will form something other than carbon dioxide and water upon combustion in air? A) C 5 H 5 N B) CH 4 C) CH OCH D) C 2 H 6 E) C 2 H 44 O 9 C 5 H 5 N contains an element other than C, H & O. Something must happen with the N, therefore something other than CO 2 and H 2 O must be formed. 4. Predict the product in the combination reaction below. Al + N 2? A) Al N 2 B) AlN C) Al 2 N D) Al N E) AlN 2 Al always forms + cations and the nitride anion is -. Therefore they must form in a 1:1 ratio: AlN 5. The formula weight of ammonium sulfate ((NH 4 ) 2 SO 4 ) is amu. A) 116 B) 100 C) 264 D) 118 E) 12 Note there are 2 ammonium ions, thus 2 N and 8 H. 2 N 2 x 14 = 28 8 H 8 x 1 = 8 1 S 1 x 2 = 2 4 O 4 x 16 = The mass % of H in methane (CH 4 ) is. A) B) 25.1 C) 7.74 D) 4.02 E) mass H 4 g mass % H = x 100% = x 100% = 25% total mass 16 g
2 7. One mole of contains the largest number of atoms. A) Al 2 (SO 4 ) B) Cl 2 C) C 10 H 8 D) S 8 E) Na PO 4 2 (Al) + (S) + 12 (O) = 17 atoms 2 (Cl) = 2 atoms 10 (C) + 8 (H) = 18 atoms 8 (S) = 8 atoms (Na) + 1 (P) + 4 (O) = 8 atoms 8. There are mol of carbon atoms in 4 mol of dimethylsulfoxide (C 2 H 6 SO). A) 6 B) 4 C) D) 2 E) 8 2 mol. C 4 mol. C2HSO 6 x = 8 mol. C 1 mol. CHSO What is the empirical formula of a compound that contains 27.0% S, 1.4% O, and 59.6% Cl by mass? A) SOCl 2 B) SOCl C) S 2 OCl D) SO 2 Cl E) ClSO 4 S: 27g / 2 g/mol ~ 1 O: 1.4 g / 16 g/mol. ~ 1 Cl: 59.6 g / 5.5 g/mol. ~ 2 therefore - SOCl 2 more accurately: S: 27g / 2 g/mol = 0.84, 0.84 / 0.84 = 1 O: 1.4 g / 16 g/mol. = 0.84, 0.84 / 0.84 = 1 Cl: 59.6 g / 5.5 g/mol. = 1.64, 1.64 / 0.84 = 2 therefore - SOCl A.82-g sample of Mg N 2 (MW=100.9 g/mol.) was combined with 7.7 g of water (MW=18.02 g/mol.) to give.60 g MgO (MW=40.0 g/mol.). What is the percent yield in the reaction? Mg N 2 + H 2 O? 2NH + MgO A) 94.5% B) 46.6% C) 78.8% D) 99.9% E) 49.4% 1 mol. Mg N mol. MgO 40. g MgO 2.82 g MgN 2 x x x = 4.58 g MgO g MgN2 1 mol. MgN2 1 mol. MgO 1 mol. H O mol. MgO 40. g MgO g H2O x x x = 17. g MgO g H2O mol. H2O 1 mol. MgO Thus Mg N 2 is the limiting reagent..60 g % yield = x 100% = 78.8% 4.58 g 11. What is the maximum mass of SO (MW=80.06 g/mol.) that can be produced by the reaction of 1.0 g of S (MW=2.07 g/mol.) with 1.0 g of O 2 (MW=2.00 g/mol.) via the equation below? 2 S (s) + O 2 (g)? 2 SO (g) A) 0.27 g B).8 g C) 1.7 g D) 2.0 g E) 2.5 g Since we start with equal masses of S and O 2, and they have the same MW, then O 2 must be the limiting reactant since the reaction requires moles O 2 per mole S. 1 mol. O2 2 mol. SO g SO 1.0 g O 2 x x x = 1.7 g SO 2.0 g O mol. O 1 mol. SO What is the solvent in an aqueous solution of sodium chloride? a) sodium b) chlorine c) water d) NaCl e) ethanol
3 1. A strong electrolyte is one that completely in solution. A) decomposes B) disappears C) ionizes D) reacts E) dissolves 14. The net ionic equation for the reaction between aqueous solutions of HCl and KOH is. A) H + + OH -? H 2 O B) HCl + KOH? H 2 O + K + + Cl - C) HCl + K + + OH -? H 2 O + KCl D) H + + Cl - + K + + OH -? H 2 O + K + + Cl - E) HCl + OH -? H 2 O + Cl - HCl + KOH? H 2 O + KCl all but H 2 O ionize, giving: H + + Cl - + K + + OH -? H 2 O + K + + Cl - however, K + and Cl - are spectator ions and are removed, leaving: H + + OH -? H 2 O 15. Which ion(s) is/are spectator ions in the reaction of BaI 2 and K 2 SO 4? A) K + only B) Ba 2+ only C) Ba 2+ and SO 4 2- D) K + and I - E) SO 4 2- and I - BaI 2 (aq) + K 2 SO 4 (aq)? BaSO 2 (s)+ 2KI (aq) all species ionize, giving: Ba I - + 2K + + SO 4 2-? BaSO 2 (s)+ 2K + + 2I - K + and I - are present on both sides of the equation and must therefore be the spectator ions. 16. The balanced net ionic equation for the reaction of aqueous solutions of Li 2 CO and CaCl 2 are mixed is A) Li 2 CO (aq) + CaCl 2 (aq)? 2LiCl (aq) + CaCO (s) B) 2Li + (aq) + 2Cl - (aq)? 2LiCl (aq) C) Li + (aq) + Cl - (aq)? LiCl (aq) D) Ca 2+ (aq) + CO 2- (aq)? CaCO (s) E) 2Li + (aq) + CO 2- (aq)? Li 2 CO (aq) Equation A is the full equation, rewritten with all aqueous species ionized gives: 2Li + (aq) + CO 2- (aq) + Ca 2+ (aq) + 2Cl - (aq)? 2Li + (aq) + 2Cl - (aq) + CaCO (s) removing the spectator ions (Li + + 2Cl - ) leaves: Ca 2+ (aq) + CO 2- (aq)? CaCO (s) 17. In which reaction does the oxidation number of oxygen increase: A) 2 SO 2 (g) + O 2 (g)? 2 SO (g) B) HCl (aq) + NaOH (aq)? NaCl (aq) + H 2 O (l) C) Ba(NO ) 2 (aq) + K 2 SO 4 (aq)? BaSO 4 (s) + 2KNO (aq) D) MgO (s) + H 2 O (l)? Mg(OH) 2 (s) E) 2 H 2 O (l)? 2 H 2 (g) + O 2 (g) O 2- + O 0? O 2- O 2-? O 2- O 2-? O 2- O 2- + O 2-? O 2- O 2-? O 0 ox. number decreased ox. number increased 18. In which species does sulfur have the highest oxidation number? A) S 8 (elemental form of sulfur) B) K 2 SO 4 C) H 2 SO D) H 2 S E) SO 2 O.N. = 0 (by def'n elements have ox. # of 0) 2(+1) + x + 4(-2) = 0, x = +6 2(+1) + x + (-2) = 0, x = +4 2(+1) + x = 0, x = -2 x + 2(-2) = 0, x = +4
4 19. The net ionic equation for the dissolution of zinc metal in aqueous hydrobromic acid is. A) 2Zn (s) + 2H + (aq)? 2Zn 2+ (aq) + H 2 (g) B) Zn (s) + 2Br - (aq)? ZnBr 2 (aq) C) Zn (s) + 2HBr (aq)? ZnBr 2 (s) + 2H + (aq) D) Zn (s) + 2HBr (aq)? ZnBr 2 (aq) + 2H + (aq) E) Zn (s) + 2H + (aq)? Zn 2+ (aq) + H 2 (g) The overall equation is: Zn (s) + 2HBr (aq)? ZnBr 2 (aq) + H 2 (g) The ionic equation is: Zn+ 2H + + 2Br -? Zn Br - + H 2 Removing the spectator ion (Br - ) leaves the net ionic eq.: Zn (s) + 2H + (aq)? Zn 2+ (aq) + H 2 (g) 20. Oxidation cannot occur without. A) water B) acid C) air D) reduction E) oxygen If one species gains electrons, another must be giving up (an equal number of) electrons 21. Which of the following is an oxidation-reduction reaction? A) H 2 CO (aq) + Ca(NO ) 2 (aq)? 2HNO (aq) + CaCO (s) B) Ba(C 2 H O 2 ) 2 (aq) + Na 2 SO 4 (aq)? BaSO 4 (s) + 2NaC 2 H O 2 (aq) C) HCl (aq) + NaOH (aq)? H 2 O (l) + NaCl (aq) D) AgNO (aq) + HCl (aq)? AgCl (s) + HNO (aq) E) Cu (s) + 2AgNO (aq)? 2Ag (s) + Cu(NO ) 2 (aq) Cu 0? Cu 2+ (oxidation) Ag +? Ag 0 (reduction) In all other cases there is no change in ox. no. Note that ions on reactant side reappear on the product side as well. 22. Which one of the following is a correct expression for molarity? A) mol solute/kg solvent B) mol solute/l solvent C) mol solute/ml solvent D) mol solute/l solution E) mol solution/l solute 2. How many grams of NaOH (MW = 40.0) are there in ml of a M NaOH solution? A) 114 B).50 x 10 C) 14.0 D).50 E) 2.19 x 10-1 L mol. NaOH 40.0 g NaOH 500 ml x x x =.50 g NaOH 1000 ml 1 L 1 mol. NaOH 24. There are mol of bromide ions in L of a 0.00 M solution of AlBr. A) B) C) D) E) Note that each mol. AlBr gives mol. Br - : AlBr? Al + + Br mol. AlBr mol. Br 1 L 1 mol. AlBr L x x = mol. Br
5 25. A M solution of will contain the highest concentration of potassium ions. A) potassium phosphate B) potassium oxide C) potassium hypochlorite D) potassium iodide E) potassium hydrogen carbonate K PO 4 (aq)? K + (aq) + PO 4 - (aq) K 2 O (aq)? 2K + (aq) + O 2- (aq) KClO (aq)? K + (aq) + ClO - (aq) KI (aq)? K + (aq) + I - (aq) KHCO (aq)? K + (aq) + HCO - (aq) 26. What mass (g) of AgBr (MW=187.8) is formed when 5.5 ml of M AgNO is treated with an excess of aqueous hydrobromic acid? A) 5.6 B) 188 C) 4.5 D) 1.2 E) 1.44 AgNO (aq) + HCl (aq)? AgBr (s) + HNO (aq) mol. AgNO 1 mol. AgBr g AgBr 1 L 1 mol. AgNO 1 mol. AgBr L x x x = 1.2 g AgBr 27. A 17.5 ml sample of a sulfuric acid (H 2 SO 4 ) solution required 29.6 ml of M NaOH for neutralization. The concentration of sulfuric acid was M. A) 4.7 B) C) D).8 E) 0.42 H 2 SO 4 (aq) + 2NaOH (aq)? Na 2 SO 4 (aq) + 2H 2 O (l) mol. NaOH 1 mol. H2SO L x x =.70 x 10 mol. HSO 1 L 2 mol. NaOH -.70 x 10 mol. H2SO 4 = M H 2 SO L 2 4
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