(a) acceptable to dry volumetric glassware at high temperature 12] 161 [7] 14] (
ID: 713617 • Letter: #
Question
(a) acceptable to dry volumetric glassware at high temperature 12] 161 [7] 14] (b) advisable to dry the interior of glassware List the rules for handling reagents and solutions 4 List the precautions that should be observed when using an Analytical Balance. 5 What is the difference between density and specific gravity of a substance? 6 Apply the Q test to the following data sets to determine whether the outlying result should be retained 3 or rejected at the 95 % confidence level: 41.27, 41.61, 41.84, 41.70 7 Generate a titration curve for the titration 25.00 mL aliquot of 0.200 M formic acid with 0.100M NaOH. Calculate the pH of the solution after the addition of 0.00; 10.00; 25.00; 40.00; 49.00, 49.90 50.00; 50.10, 55.00 and 60mL of the base. Prepare a titration curve from the data on GRAPH PAPER or on Excel. [20] 8 What mass of solid PbC12 (278.10 g/mol) is formed when 200 mL of 0.125 M Pb2+ are mixed with 400 mL of 0.175 M CIExplanation / Answer
8) Write down the balanced chemical equation for the formation of PbCl2 as
Pb2+ (aq) + 2 Cl- (aq) ---------> PbCl2 (s)
As per the stoichiometric equation,
1 mole Pb2+ = 2 mole Cl- = 1 mole PbCl2.
Millimoles of Pb2+ added = (200 mL)*(0.125 M) = 25.0 mmole.
Millimoles of Cl- added = (400 mL)*(0.175 M) =70.0 mmole.
Determine the limiting reactant.
Pb2+: (25.0 mmole Pb2+)*(2 mole Cl-/1 mole Pb2+) = 50.0 mmole.
Cl-: (70.0 mmole Cl-)*(1 mole Pb2+/2 mole Cl-) = 35.0 mmole.
Clearly, we do not have 35.0 mmole Pb2+ but we have excess Cl- (than required to react with 25.0 mmole Pb2+). Hence, Pb2+ is the limiting reactant.
Millimoles of PbCl2 formed = (25.0 mmole Pb2+)*(1 mole PbCl2/1 mole Pb2+) = 25.0 mmole PbCl2.
Mass of PbCl2 produced = (25.0 mmole)*(1 mole/1000 mmole)*(278.10 g/mol) = 6.9525 g (ans).
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