A 185.0 mL sample of 1.200 M Pb(NO₃)₂ is mixed with 52.50 mL of 1.500 M NaCl, and the PbCl₂ precipitate is filtered from the solution. Then 200.0 mL of 3.000 M NaBr is added to the remaining solution, and the PbBr₂ precipitate is also collected and dried. What is the mass (in grams) of the PbBr₂ precipitate, assuming the yield in each precipitation step is 100%?
A 185.0 mL sample of 1.200 M Pb(NO₃)₂ is mixed with 52.50 mL of 1.500 M NaCl, and the PbCl₂ precipitate is filtered from the solution. Then 200.0 mL of 3.000 M NaBr is added to the remaining solution, and the PbBr₂ precipitate is also collected and dried. What is the mass (in grams) of the PbBr₂ precipitate, assuming the yield in each precipitation step is 100%?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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A 185.0 mL sample of 1.200 M Pb(NO₃)₂ is mixed with 52.50 mL of 1.500 M NaCl, and the PbCl₂ precipitate is filtered from the solution. Then 200.0 mL of 3.000 M NaBr is added to the remaining solution, and the PbBr₂ precipitate is also collected and dried. What is the mass (in grams) of the PbBr₂ precipitate, assuming the yield in each precipitation step is 100%?
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Step 1: Defining double displacement reaction
VIEWStep 2: Calculation for the moles of Lead nitrate, NaCl and NaBr
VIEWStep 3: Calculation for the moles of lead nitrate remains in solution after the addition of NaCl
VIEWStep 4: Finding out the limiting reagent
VIEWStep 5: Finding out the moles of lead bromide forming
VIEWStep 6: Calculation for mass of lead bromide
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