What volume of 2.0 M hydrochloric acid is needed to completely react with the amount of calcium carbonate in Part 2a above? c. Based on Parts 2a and 2b above, how many moles of water would be produced? 3. Ammonium chloride and calcium hydroxide react according to the following balanced equation: 2 NH4Cl(aq) + Ca(OH)2(aq) ⋄ CaCl2(aq) + 2 NH3(g) + 2 H2O(l) a. What mass of ammonium chloride is needed to make 3.0 liters of a 1.5 M ammonium chloride solution?
If 0.100 mol of hydrogen iodide is placed in a 1.0 L container and allowed to reach equilibrium, find the concentrations of all reactants and products at equilibrium. 2 HI (g) === H2 (g) + I2 (g) Ke = 1.84(10-2 [H2]=[I2]= 1.07(10-2 mol/L, [HI]=7.86(10-2 mol/L 6. A 1.00 L reaction vessel initially contains 9.28(10-3 moles of H2S. At equilibrium, the concentration of H2S of 7.06(10-3 mol/L. Calculate the value of Ke for this system.
3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate.
Investigating the Chemical Composition of Alka Selzter Problem: How much sodium bicarbonate is in a tablet of Alka Seltzer? Purpose: To discover the amount of sodium bicarbonate, in grams, in a one tablet of Alka Seltzer. Materials: one Alka Seltzer tablet 50 mL of vinegar post-it note Triple beam balance beaker Procedure: 1) Put the post-it on the triple beam balance and zero the balance. 2) Measure the masses of the alka-seltzer tablet, and the beaker + vinegar. Record them.
Steam Distillation: The isolation of limonene 1. Purpose The purpose of the experiment is to be familiar with the use of steam distillation, to learn how to extract an organic solvent and remove a volatile liquid by a rotary evaporator, and to determine the purity of the limonene. Steam distillation was used because it is a special type of distillation especially for temperature sensitive materials. Dichloromethane was used to extract the aqueous mixture. Magnesium sulfate was used to absorb water and simple filtration was used to remove magnesium sulfate.
The reaction is a synthesis. The Kc for this reaction is Kc = 49.7 at 458oC [Answer: [H2] = [I2] = 1.2 × 10-2 M, [HI] = 8.6 × 10-2 M] 2. Iodine and bromine react to give iodine monobromide, IBr. What is the equilibrium composition of a mixture at 150oC that initially contained 0.0015 mol each of iodine and bromine in a 5.0 L vessel? The equilibrium constant Kc for this reaction at 150oC is 1.2 102.
Chlorine gas can be produced in the laboratory by adding concentrated hydrochloric acid to manganese(IV) oxide in the following reaction: MnO2(s) + 4HCl(aq) ( MnCl2(aq) + 2H2O(l) + Cl2(g) a. Calculate the mass of MnO2 needed to produce 25.0 g Of Cl2 ans: 30.7 g MnO2 b. What mass of MnCl2 is produced when 0.091 g of C12 is generated? ans: 0.16 g MnCl2 1. How many moles of ammonium sulfate can be made from the reaction of 30.0 mol of NH3 with H2SO4 according to the following equation: ans.
| Observations of Chemicals | Zinc Sulfate | Powder of a white solid | Barium Iodide | Powder of a white solid. | Deionized water | Liquid, transparent. | Trial # | BaI2 | ZnSO4 | Theoretical Yield of ZnI2 | Actual Yield | Percent Yield | 1 | .67g | .45g | .499820g | .52g | 104% | 2 | .67g | .45g | .499820g | .52g | 104% | 3 | .66g | .46g | .493117g | .48g | 97% | Calculations for Cost | Double Replacement | Synthesis | 0.48 grams of Zinc Sulfate - $0.02 | 1.00 gram Granular Zinc - $62.50 | 0.67grams of Barium Iodine Dihydrate - $0.886 | 2.00 gram Iodine - ($74.90 × 2) - $149.80 | 0.52 grams of Zinc Iodide - $0.906 | 1.00 gram zinc - $0.212 | 1000 grams of Zinc Iodide = $1,923.00 | 1000 grams of Zinc Ioidide = $212.30 | Focus Question Should chemists prepare Zinc Iodide, from its Elements or from a Double Replacement Reaction between Barium Iodide and Zinc Sulfate?