20. mol H2 reacts with 8.0 mol O2 to produce H2O. Determine the number of grams reactant in excess and number of grams H2O produced. Identify the limiting reactant. 8.1 g H2 , 2.9 x 102 g H2O 17. How many litres of O2 gas are required to produce 100. g Al2O3?
25.0 cm3 of a 0.10 moldm-3 solution of sodium hydroxide was titrated against a solution of hydrochloric acid of unknown concentration. 27.3 cm3 of the acid was required. What was the concentration of the acid? 3. 25 cm3 of a solution of sodium hydroxide reacts with 15 cm3 of 0.1 mol/dm3 HCl.
Aim : a) To determine reduction potentials of several redox couples. b) To determine the effect of concentration changes on cell potential. c) To determine the molar concentration of Cu2+ in the unknown using Nernst equation. Procedure : Please refer to the laboratory manual page 77 – 79. Results : Reduction Potentials of Several Redox Couples Galvanic Cell Measured Ecell Anode Equation for Anode Reaction Cathode Equation for Cathode Reaction Cu-Zn +1.19 V Zn Zn → Zn2+ + 2e- Cu Cu2+ + 2e- → Cu Cu-Mg +3.23 V Mg Mg → Mg2+ + 2e- Cu Cu2+ + 2e- → Cu Cu-Fe +0.89 V Fe Fe → Fe2+ + 2e- Cu Cu2+ + 2e- → Cu Zn-Mg +2.00 V Mg Mg → Mg2+ + 2e- Zn Zn2+ + 2e- → Zn Fe-Mg +1.54 V Mg Mg → Mg2+ + 2e Fe Fe2+ + 2e → Fe Zn-Fe +0.29 V Zn Zn → Zn2+ + 2e- Fe Fe2+ + 2e- → Fe Balanced net reaction Cu-Zn Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s) Cu-Mg Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s) Cu-Fe Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s) Zn-Mg Mg (s) + Zn2+ (aq) → Mg2+ (aq) + Zn (s) Fe-Mg Mg (s) + Fe2+ (aq) → Mg2+ (aq) + Fe (s) Zn-Fe Zn (s) + Fe2+ (aq) → Zn2+ (aq) + Fe (s) Zn-Mg = + 2.00 V, Mg-Cu = + 3.23 V The cell potential of Zn-Cu = 3.23 V – 2.00 V = 1.23 V The measured cell potential of Zn-Cu = +1.19 V The value of the sum of the Zn-Mg and Zn-Cu cell potentials are nearly the same as the Cu-Mg cell potential.
Percent H2O in Hydrate is equal 0.34/2.33=14.6% 3. The general formula of barium chloride hydrate is BaClg-nHZO, where n is the number of water molecules. Calculate the theoretical percent water for each value of n—divide the sum of the atomic masses due to the water molecules by the sum of all the atomic masses in the hydrate, and multiply the result by 100. Complete the table. | BaCl2 | BaCl2•H2O | BaCl2•2H2O | BaCl•3H2O | Sum of atomic masses (BaCl2) | 208.23 | 208.23 | 208.23 | 208.23 | Sum of atomic masses (nH2O) | 0 | 18.02 | 36.04 | 54.06 | Sum of atomic masses (hydrate) | 208.23 | 226.25 | 244.27 | 262.29 | Percent water in hydrate (theoretical) | 0% | 7.96% | 14.75% | 20.61% | In this lab we used a Balance, centigram
Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium. The reaction is as follows: Mg + 2HCl -> H2 + Mg2+ (aq) + 2Cl- (aq) There is a one to one relationship between the number of moles of hydrogen gas evolved and the
n (3) Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride. co NaOH(s) → Na+(aq) + OH–(aq) ∆H1 = ? Chemistry with Vernier py In this experiment, you will use a Styrofoam-cup calorimeter to measure the heat released by three reactions. One of the reactions is the same as the combination of the other two reactions. Therefore, according to Hess’s law, the heat of reaction of the one reaction should be equal to the sum of the heats of reaction for the other two.
Empirical formula: CH5N Steps for molecular formula: 1- Calculate the molar mass of the empirical formula. 2- Divide the known (given) molar mass by the calculated empirical formula molar mass to get a whole number 3- Multiply that whole number through subscripts of the empirical formula to obtain the molecular formula. Example CH5N 12.01 g C x 1 C= 12.01 g/mol 1.008 g H x 5 H = 5.040
Practice Question for Exam 2 Chem 103 |A chemical reaction requires 30.77 kJ. How many kilocalories does this correspond to? | |A) 7,354 kcal B) 7.354 kcal C) 128.7 kcal D) 0.1287 kcal E) 30.77 kcal | |776 J is the same quantity of energy as | |A) 7.76 × 105 kJ. B) 1.85 × 102 kcal. C) 0.185 kcal.
Percent Yield Calculations 1) Balance this equation and state which of the six types of reaction is taking place: ____ Mg + ____ HNO3 ( ____ Mg(NO3)2 + ____ H2 Type of reaction: __________________________ 2) If I start this reaction with 40 grams of magnesium and an excess of nitric acid, how many grams of hydrogen gas will I produce? 3) If 1.7 grams of hydrogen is actually produced, what was my percent yield of hydrogen? 4) Balance this equation and state what type of reaction is taking place: ____ NaHCO3 ( ____ NaOH + ____ CO2 Type of reaction: __________________________ 5) If 25 grams of carbon dioxide gas is produced in this reaction, how many grams of sodium hydroxide should be produced? 6) If 50 grams of sodium hydroxide are actually produced, what was my percent yield?
Using the G° data in your Appendix B, calculate the change in Gibbs free energy for each of the following reactions. In each case, indicate whether the reaction is spontaneous under standard conditions. a) H2 (g) + Cl2 (g) → 2HCl (g) b) MgCl2 (s) + H2O (l) → MgO (s) + 2 HCl (g) c) 2 NH3 (g) → N2H4 (g) + H2 (g) d) 2 NOCl (g) → 2 NO (g) + Cl2 (g) 4. From the values given for ΔH° and ΔS°, calculate ΔG° at 25°C for each of the following reactions. If the reaction is not spontaneous under standard conditions at 298K, at what temperature (if any) would the reaction become spontaneous?