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	Solutions to Lab Problems
	1) A 157.82 g sample of MgNO₃ ^. ???H₂O in an evaporating dish is heated until it is dry. The mass of the evaporating dish is 152 g. The mass of the anhydrous sample and the evaporating dish is 154.58 g MgNO₃. What is the formula for the hydrate and name the hydrate? sample + dish = 157.82 g -dish = 152 g 5.82 g sample w water dry sample + dish = 154.58 g -dish = 152 g 2.58 g sample w/out water wet sample 5.82 g - dry sample 2.58 g water 3.24 g [2.58 g MgNO₃ ][1 mol MgNO₃ ] = .03 mol [86 g MgNO₃ ] [3.24 g H₂O][1 mol H₂O] = 0.18 mol [18 g H₂O] [.03] = 1 [.18] = 6 [.03] [.03] MgNO₃ ^. 6H₂O magnesium nitrate hexahydrate 2) A 0.36 g sample of Mg is reacted with an excess of hydrochloric acid at 24C and 100.1 KPa. The resulting hydrogen gas is collected under water. The objective is to generate the ideal gas constant experimentally. The volume of the gas collected was 368 mL. What was the ideal gas constant found in lab? What was the percent error? Mg + 2HCl --> MgCl₂ + H₂ [0.36 g Mg][ 1 mol Mg][1 mol H₂] = 0.015 mol H₂ [ 24.3 g Mg][1 mol Mg] mol = 0.015 mol H₂atmospheric pressure = 100.1 KPa water vapor pressure at 24C is 3.0 KPa pressure of dry gas = 97.1 KPa temperature = 24 + 273K = 297 K volume = 368 mL = 0.368 L = 0.368 dm³ pv = nRt R= pv nt R = [97.1 KPa][0.368 dm^3] = 8.02 Kpa dm³ [0.015 mol][297K] mol K % error = [8.31-8.02] * 100 = 2.3% error [8.31] 3) How many mL of 1.00 M NaOH are needed to neutralize a 10.00 mL aliquot of 1 M HCl after it has been partially neutralized to a .20 M HCl solution by the addition of 0.25 g CaCO₃ antacid? HCl + NaOH <--> NaCl + HOH [10.00 mL HCl][ 1 L HCl][.20 mol HCl][1 mol NaOH][1L NaOH ][1000 mL NaOH] = [1000mL HCl][1 L HCl ][1 mol HCl ][1.0 mol NaOH][1L NaOH ] 10 mL NaOH required to neutralize HCl