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What quantity in moles of precipitate will be formed when 80.0 mL of 2.00 M Nal is reacted with 100.0 mL of 0.900 M Pb(NO₂)₂ in the following chemical reaction? 2 Nal (aq) + Pb(NO₂)₂ (aq) → Pbl₂ (s) + 2 NaNO₂ (aq)

What quantity in moles of precipitate will be formed when 80.0 mL of 2.00 M Nal is reacted with 100.0 mL of 0.900 M Pb(NO₂)₂ in the following chemical reaction? 2 Nal (aq) + Pb(NO₂)₂ (aq) → Pbl₂ (s) + 2 NaNO₂ (aq)

Chemistry
Chemistry
10th Edition
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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**Chemical Reaction Problem** **Problem Statement:** What quantity in moles of precipitate will be formed when 80.0 mL of 2.00 M NaI is reacted with 100.0 mL of 0.900 M Pb(NO₃)₂ in the following chemical reaction? **Chemical Equation:** \[ 2 \, \text{NaI (aq)} + \text{Pb(NO₃)₂ (aq)} \rightarrow \text{PbI₂ (s)} + 2 \, \text{NaNO₃ (aq)} \] **Explanation:** In this reaction: - Sodium iodide (NaI) is combined with lead(II) nitrate (Pb(NO₃)₂) to produce lead(II) iodide (PbI₂), a solid precipitate, and sodium nitrate (NaNO₃), which remains in solution. - The balanced equation indicates that two moles of NaI react with one mole of Pb(NO₃)₂ to form one mole of PbI₂. **Quantitative Analysis:** Given: - Volume of NaI solution = 80.0 mL = 0.0800 L - Molarity of NaI solution = 2.00 M - Volume of Pb(NO₃)₂ solution = 100.0 mL = 0.100 L - Molarity of Pb(NO₃)₂ solution = 0.900 M Calculate the moles of each reactant and determine which is the limiting reactant. This will allow you to find the moles of PbI₂ precipitate formed.
Transcribed Image Text:**Chemical Reaction Problem** **Problem Statement:** What quantity in moles of precipitate will be formed when 80.0 mL of 2.00 M NaI is reacted with 100.0 mL of 0.900 M Pb(NO₃)₂ in the following chemical reaction? **Chemical Equation:** \[ 2 \, \text{NaI (aq)} + \text{Pb(NO₃)₂ (aq)} \rightarrow \text{PbI₂ (s)} + 2 \, \text{NaNO₃ (aq)} \] **Explanation:** In this reaction: - Sodium iodide (NaI) is combined with lead(II) nitrate (Pb(NO₃)₂) to produce lead(II) iodide (PbI₂), a solid precipitate, and sodium nitrate (NaNO₃), which remains in solution. - The balanced equation indicates that two moles of NaI react with one mole of Pb(NO₃)₂ to form one mole of PbI₂. **Quantitative Analysis:** Given: - Volume of NaI solution = 80.0 mL = 0.0800 L - Molarity of NaI solution = 2.00 M - Volume of Pb(NO₃)₂ solution = 100.0 mL = 0.100 L - Molarity of Pb(NO₃)₂ solution = 0.900 M Calculate the moles of each reactant and determine which is the limiting reactant. This will allow you to find the moles of PbI₂ precipitate formed.
**What is the difference between actual yield and theoretical yield?** A) Actual yield has to do only with the reactants of a reaction, and theoretical yield has to do only with the products of a reaction. B) Actual yield is how much is actually produced in a reaction (it has to be given or measured), and theoretical yield is a calculation that has to be done. C) Actual yield is how much the reaction produces, and theoretical yield is how much a reaction consumes. D) Actual yield has to do with how much you can actually get out of a reaction, and theoretical yield is how much you can get out of a reaction if you have maximum starting materials.
Transcribed Image Text:**What is the difference between actual yield and theoretical yield?** A) Actual yield has to do only with the reactants of a reaction, and theoretical yield has to do only with the products of a reaction. B) Actual yield is how much is actually produced in a reaction (it has to be given or measured), and theoretical yield is a calculation that has to be done. C) Actual yield is how much the reaction produces, and theoretical yield is how much a reaction consumes. D) Actual yield has to do with how much you can actually get out of a reaction, and theoretical yield is how much you can get out of a reaction if you have maximum starting materials.
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