Thank you for visiting A 99 8 mL sample of a solution that is 12 0 KI by mass density 1 093 g mL is added to 96 7. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Answer:
[tex]m_{PbI_2}=18.2gPbI_2[/tex]
Explanation:
Hello,
In this case, we write the reaction again:
[tex]Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)[/tex]
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
[tex]n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI[/tex]
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
[tex]0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI[/tex]
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
[tex]m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2[/tex]
Best regards.
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Rewritten by : Jeany
Answer:
Mass PbI2 = 18.19 grams
Explanation:
Step 1: Data given
Volume solution = 99.8 mL = 0.0998 L
mass % KI = 12.0 %
Density = 1.093 g/mL
Volume of the other solution = 96.7 mL = 0.967 L
mass % of Pb(NO3)2 = 14.0 %
Density = 1.134 g/mL
Step 2: The balanced equation
Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)
Step 3: Calculate mass
Mass = density * volume
Mass KI solution = 1.093 g/mL * 99.8 mL
Mass KI solution = 109.08 grams
Mass KI solution = 109.08 grams *0.12 = 13.09 grams
Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL
Mass of Pb(NO3)2 solution = 109.66 grams
Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams
Step 4: Calculate moles
Moles = mass / molar mass
Moles KI = 13.09 grams / 166.0 g/mol
Moles KI = 0.0789 moles
Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol
Moles Pb(NO3)2 = 0.0463 moles
Step 5: Calculate the limiting reactant
For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3
Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles
Step 6: Calculate moles PbI2
For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3
For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2
Step 7: Calculate mass of PbI2
Mass PbI2 = moles PbI2 * molar mass PbI2
Mass PbI2 = 0.03945 moles * 461.01 g/mol
Mass PbI2 = 18.19 grams
