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1. 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 mL of another solution that is 14.0% Pb(NO3)2 by mass (density: 1.134 g/mL). How many grams of PbI2 should form?

\[ \text{Reaction: } \text{Pb(NO}_3\text{)}_2\text{(aq) + 2 KI(aq) } \rightarrow \text{PbI}_2\text{(s) + 2 KNO}_3\text{(aq)} \]

Answer :

Answer:

The mass of PbI2 will be 18.2 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 = 18.2 grams

Final answer:

To find the mass of PbI2 that forms, we need to first find the mass of KI and Pb(NO3)2 in the solutions, change these into moles and then use the stoichiometry of the balanced chemical equation to determine the moles of PbI2 that would form. Converting these moles of PbI2 back into grams gives us approximately 18.44 grams of PbI2.

Explanation:

  • Firstly, we need to determine the mass of both solutions. Using the formulas density = mass/volume and mass = density * volume, we find that the mass of the KI solution is 109.16 grams and the mass of the Pb(NO3)2 solution is 109.66 grams.
  • Next, knowing that 12.0% of the KI solution is KI and 14.0% of the Pb(NO3)2 solution is Pb(NO3)2, we can determine the mass of KI and Pb(NO3)2 in the solutions - 13.09 grams of KI and 15.35 grams of Pb(NO3)2.
  • From the balanced chemical equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KI to form PbI2. Converting the grams of KI and Pb(NO3)2 to moles (using the molar mass of each compound), we get 0.077 moles of KI and 0.040 moles of Pb(NO3)2.
  • However, according to the stoichiometry of the reaction, the mole ratio of Pb(NO3)2 to KI is 1:2. So the limiting reactant is Pb(NO3)2. So, in the reaction, only 0.040 moles of PbI2 would be formed.
  • Finally, to find out the answer, we convert these moles of PbI2 back into grams using the molar mass of PbI2 (461.01 g/mol), which gives us approximately 18.44 grams of PbI2.

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