CHM 134 General Chemistry I Exam 2 Review, Dr. Steel. 1. Give the oxidation number of sulfur in each of these compounds.

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CHM 1 General Chemistry I Exam Review, Dr. Steel Name 1. Give the oxidation number of sulfur in each of these compounds. H S SO H SO SO -. In the lab you reacted magnesium metal and oxygen gas to produce solid magnesium oxide. a. Write the balanced equation for the reaction, including phase notation. b. How much magnesium oxide should be produced if you begin with.71 g of magnesium? c. What mass of oxygen gas would you need to produce.7 grams of magnesium oxide if the reaction only had a 7.6% yield?. Indicate whether each species is soluble or insoluble in water. NiS Cu(OH KOH Ca (PO (NH SO Cu(ClO. Determine the masses of calcium fluoride produced and excess reactant remaining when 16.8 grams of calcium nitrate react with 17.50 grams of ammonium fluoride. Ca(NO (s + NH F(s CaF (s + N O(g + H O(g 5. Indicate the oxidizing and reducing agent in this reaction and use oxidation numbers to balance the equation. K Cr O 7 (aq + HI(aq KI(aq + CrI (aq + I (s + H O(l

6. Will each these reactions occur or not? NaCl(aq + LiBr(aq (Both products are soluble salts H PO (aq + CaCl (aq (Ca (PO is insoluble sodium iodide + lead (II nitrate (PbI is insoluble Ni(s + MgCl (aq (Nickel can not replace magnesium CH OH(aq + NaOH(aq (water is produced hydrochloric acid + sodium oxide (water is produced ZnSO (aq + Cu(s (copper can not replace zinc 7. In the lab you use a 100.0 ml volumetric flask to prepare a solution of potassium carbonate (K. You mass 1.7 grams of K into the flask and fill it with distilled water. a. What is the concentration of the solution? b. What volume of the potassium carbonate solution do you need to prepare 50.00 ml of 0.15 M K solution? 8. In the lab you prepare 150.0 ml of.679 M sulfuric acid and add 11.76 grams of aluminum metal to it. The reaction is shown below. Al(s + H SO (aq Al (SO (aq + H (g a. What mass of hydrogen gas is produced by the reaction? b. How many moles of the excess reactant remain after the reaction?

CHM 1 General Chemistry I Name SOLUTIONS Exam Review, Dr. Steel 1. Give the oxidation number of sulfur in each of these compounds. - H S + SO +6 H SO + SO -. In the lab you reacted magnesium metal and oxygen gas to produce solid magnesium oxide. a. Write the balanced equation for the reaction, including phase notation. b. How much magnesium oxide should be produced if you begin with.71 g of magnesium? c. What mass of oxygen gas would you need to produce.7 grams of magnesium oxide if the reaction only had a 7.6% yield? A. Mg(s + O (g MgO(s 1mol Mg mol Mg 0.1g MgO B..71 g Mg =.55 g MgO.1g Mg mol MgO 1mol MgO C. Use the 7.6% yield to calculate the theoretical yield of MgO:.7 grams MgO = 0.76(x, where x = theoretical yield of MgO.7 g MgO x = =.99 g MgO (theoreticalyield 0.76 Now determine the amount of oxygen required for this theoretical yield: 1mol MgO 1mol O.00 g O.99 g MgO = 0.1g MgO mol MgO 1mol O 1.98 g O. Indicate whether each species is soluble or insoluble in water. INSOLUBLE NiS INSOLUBLE Cu(OH SOLUBLE KOH INSOLUBLE Ca (PO SOLUBLE (NH SO SOLUBLE Cu(ClO

. Determine the masses of calcium fluoride produced and excess reactant remaining when 16.8 grams of calcium nitrate react with 17.50 grams of ammonium fluoride. Ca(NO (s + NH F(s CaF (s + N O(g + H O(g 1mol NH F 1mol CaF 78.08 g CaF 17.50 g NH F 7.0 g NH F mol NH F 1mol CaF = 18. g CaF 16.8 g Ca(NO 1mol Ca(NO 16.10 g Ca(NO 1mol CaF 1mol Ca(NO 78.08 g CaF 1mol CaF = 7.99 g CaF 16.8 g Ca(NO 1mol Ca(NO 16.10 g Ca(NO mol NH F 1mol Ca(NO 7.0 g NH F = 7.58 g NH 1mol NH F F 17.50 g NH F available 7.58 g NH F used = 9.9 g NH F leftover 5. Indicate the oxidizing and reducing agent in this reaction and use oxidation numbers to balance the equation. Reducing Agent: -1 (oxidized 1e=e 0 I -1 in KI 1 K Cr O 7 (aq + 1 HI(aq KI(aq + CrI (aq + I (s + 7 H O(l Oxidizing Agent: +6 (reduced e=6e + Cr +6 in K Cr O 7 6. Will each these reactions occur or not? (You may write the products down to help you, but are not required to do so. NO NaCl(aq + LiBr(aq (Both products are soluble salts YES H PO (aq + CaCl (aq (Ca (PO is insoluble YES sodium iodide + lead (II nitrate (PbI is insoluble NO Ni(s + MgCl (aq (Nickel can not replace magnesium YES CH OH(aq + NaOH(aq (water is produced YES hydrochloric acid + sodium oxide (water is produced NO ZnSO (aq + Cu(s (copper can not replace zinc

7. In the lab you use a 100.0 ml volumetric flask to prepare a solution of potassium carbonate (K. You mass 1.7 grams of K into the flask and fill it with distilled water. a. What is the concentration of the solution? 1.7 g K 100.0 ml Molarity : M 1mol K 18.1g K 1L 1000 ml moles = = V = 0.1000 L = 0.1066 mol K 0.1000 L 0.1066 mol K = 1.066 M K b. What volume of the potassium carbonate solution do you need to prepare 50.00 ml of 0.15 M K solution? M1V1 = M V (volume will be in L, so 50.00 ml = 0.05000 L M V (0.15 M(0.05000 L V1 = = = 0.00586 L = 5.86 ml M (1.066 M 1 8. In the lab you prepare 150.0 ml of.679 M sulfuric acid and add 11.76 grams of aluminum metal to it. The reaction is shown below. Al(s + H SO (aq Al (SO (aq + H (g a. What mass of hydrogen gas is produced by the reaction? b. How many moles of the excess reactant remain after the reaction? Determine the limiting reactant: 1mol Al mol H SO 1L H SO 11.76 g Al = 0.197 L H SO = 19.7 ml H SO 6.98 g Al mol Al.679 mol H SO You have 150.0 ml H SO, therefore it is in excess and aluminum is the limiting reactant. Calculate the mass of hydrogen produced: 1mol Al mol H.016 g H 11.76 g Al = 1.18 g H 6.98 g Al mol Al 1mol H We use 19.7 ml of the H SO (from the first calculation. Therefore the excess is: 150.0 ml 19.7 ml = 10. ml H SO leftover. So, the moles are found: moles = M V = (.670 M(0.010 L = 0.08 mol H SO