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?
Experimental Data Trial 1 Trial 2 Trial 3 Sample Volume 10 mL 10 mL 10 mL Weight sample + capped vial 26.5g 26.5g 26.3g Weight vial + cap 15.5g 15.4g 15.4g Sample weight 11.0g 11.1g 10.9g Calculated Results Trial 1 Trial 2 Trial 3 Density of samples: 1.10g/mL 1.11g/mL 1.09g/mL Average density: 1.10g/mL 1.10g/mL 1.10g/mL Deviation from Average: ±0.00g/mL ±0.01g/mL -0.01g/mL Average Deviation: add absolute values of deviations from average and divide by the number of trials. +0.00++0.01+-0.01= 0.02 ∴ Therefore, the average deviation in density = ±0.02/3 = ±0.01 Average weight of 10mL Volume Samples 10mL x 1.1g/mL = 11g Percent Inherent Error in Average Weight of Samples ±0.1 g/reading*2 readings 11* 100 = 1.8~2% Percent Inherent Error in Average Volume of Samples ±0.1 mL/reading*1 readings 10 mL*100 = 1% Total Percent Inherent Error in Density ±2%+ ±1%= ±3% Inherent Error in Density 1.10g/mL x (±3/100) = ±0.03g/mL Conclusions Since the deviations from the average density are equal to or less than the total possible inherent error computed, it is concluded that the precision of the data collected was good. Second
1. Be sure the data table is viewable. 2. Right-click (PC) or Command-Click (Mac) on the table and select print. Part I: Density of Unknown Liquid | | Trial 1 | Trial 2 | Trial 3 | Mass of Empty 10 mL graduated cylinder (grams) | 2.56g | 2.55g | 2.60g | Volume of liquid (milliliters) | 8.50mL | 8.80mL | 8.90mL | Mass of graduated cylinder and liquid (grams) | 36.10g/mL | 36.60g/mL | 36.90g/mL | Part II: Density of Irregular-Shaped Solid | Mass of solid (grams) | 41.000g | 38.890g | 42.960g | Volume of water (milliliters) | 50mL | 55.5mL | 53.1mL | Volume of water and solid (milliliters) | 54.9g/mL | 60g/mL | 58.1g/mL | Part III: Density of Regular-Shaped Solid | Mass of solid (grams) | 28.00g | 27.70g | 28.10g | Length of solid (centimeters) | 5.20cm | 5.00cm | 4.50cm | Width of solid (centimeters) | 3.00cm | 4.00cm | 3.50cm | Height of solid (centimeters) | 2.50cm | 3.00cm | 2.00cm | Calculations Show all of your work for each of the following calculations and be careful to follow significant figure rules in each calculation.
0.00512g ZnI2/mL of solution 0.00512g/319.18 g/mol=1.6*10-5 mol 1.6*10-5 mol/(1*10-3L)=0.016M c. 0.00806 moles of ZnI2/500 mL of solution 0.00806mol/(500*10-3)L=0.016M d. 0.0161 moles of ZnI2/L of solution 0.0161mol/1L=0.016M Exercise 4: a. The moles of ZnI2: 0.25M*(250*10-3)L=0.0625mol b. 0.25M*(250*10-3)L=0.0625mol The mass of ZnI2: 0.0625mol*319.18 g/mol=19.95g c. 0.25M*(500*10-3)L=0.125mol 0.125mol*319.18 g/mol=39.9g ZnI2 d. 0.0125mol/0.25M=0.05L Exercise 5: a. 0.125M*(100*10-3)L=0.0125mol b. 0.0625mol/0.125M=0.5L=500mL Calculation for preparing the EDTA solution Exercise 6 a.
In the first part, five 100 mL flasks of 5 mL ligand solution, 5 mL 2 M sodium acetate, 4 mL 3 M NH2OH, and 1-5 mL Fe2+ solution are diluted with water. The absorption spectrum for varying concentrations of Fe2+ are measured using a spectrophotometer and the data is graphed in Excel. The slope of the line is ε in the Beer-Lambart equation A = εcl. In the second part of the experiment, eleven flasks containing diluted stock solutions of Fe2+ and ligand are mixed with 5 mL 2 M sodium acetate and 4 mL 3 M NH2OH and diluted with water. The absorption spectrum is measured using a spectrophotometer and the data is graphed in Excel.
c. Prepare the solution by dissolving 38.90 grams of ZnI2 with 500 mL of water. d. 0.0125/0.25 = 0.05 L = 50 mL. This produces 0.0125 moles of ZnI2 5. Exercise 5: a. (0.125)(0.1) = 0.0125 moles of solute b. Pour 50 mL of the stock solution to get the number of moles needed.
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.
There will have some error. 2) A volatile liquid was allowed to evaporate in a 43.298 g flask that has a total volume of 252 ml. the temperature of the water bath was 100˚C at the atmospheric pressure of 776 torr. The mass of the flask and condensed vapor was 44.173 g. calculate the molar mass of the liquid. T = 273 + 100 = 373 V = 252 mL = 1 L / 1000 mL = 0.252 L P = 776 Torr R= 0.0821 mass of 44.173 - 43.298 g = 0.875g moles of gas = PV / RT = 776 x .252 / 62.363 x (273+100) =0.00841 moles molar mass = 0.875g / 0.00841 moles = 104.1 g/
- 0.5 M: Add 6.25 mL of the 2M HCl and mix with distilled water until the entire solution is measured to be about 25 mL Length and Mass of Magnesium Ribbon: • Original length: 30 cm • Original mass: 0.35 g • Length of ribbon used for experiment: 1.5 cm • Mass of ribbon used for experiment: 0.0175 g Concentration and Time data: 2.0 M 1.5 M 1.0 M 0.75 M 0.5 M Time (s) #1 13 s 18 s 38 s 81 s 225 s Time (s) #2 11 s 17 s 39 s 87 s 418 s Concentration and Temperature data: 2.0 M 1.5 M 1.0 M 0.75 M 0.5 M Temperature (Celsius) #1 30.1 30.1 30.2 30.5 28.5 Temperature (Celsius) #2 30.8 30.4 30.5 30.3 29.1 Results Table: Rate of Reaction for each concentration: 2.0 M 1.5 M 1.0 M 0.75 M 0.5 M Rate of Reaction 0.00145 0.001 0.000454 0.000208 0.0000755 Plot of Concentration vs Rate of
A. 0.420 M B. 0.567 M C. 0.042 M D. 0.325 M _____ 3. What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose, C6H12O6, per 100.0 g of H2O ? (Kf for water = 1.86 °C /m) A.