# Homework 54

- Page ID
- 28968

9.33, 10.60

## 9.33

Using this precipitation reaction:

2KI (aq) + Pb(NO_{3})_{2 }(aq)_{ }----> PbI_{2} (s)_{ }+ 2KNO_{3} (aq) *Hint: Potassium Iodine and Lead (II) Nitrate reaction

Find the volume of 0.155 M of KI needed to completely react with 0.0754 L of 0.108 M of Pb(NO_{3})_{2}.

__Definitions:__

Precipitation Reaction: The mixing of two reacting solutions in which a solid forms as a result.

__Helpful Equations:__

\[Molarity = \dfrac{moles}{Liters}\]

__Solving Strategy:__

Step 1: Establish the ratio of moles of KI to moles of Pb(NO_{3})_{2 }from the given balanced equation._{ }

*Hint: Identify the coefficients of the two reagents

Step 2: We have been given the molarity and volume for Pb(NO_{3})_{2. }We can now calculate the actual number of moles of Pb(NO_{3})_{2 }in this reaction using the Molarity equation.

Step 3: Use the number of moles of Pb(NO_{3})_{2 }calculated in Step 2 to now find moles of KI. Do this using dimensional analysis. Make sure to use the mole to mole ratio found in Step 1 for the conversion.

Step 4: Finally, plug the calculated number of moles of KI found in Step 3 to solve for the volume of KI needed. Since the molarity has been given and the number of moles has been calculated, we can plug these values into the Molarity equation once again.

__Solution:__

Step 1: 2 mole of KI : 1 mole of Pb(NO_{3})_{2}

Step 2: \[0.108M=\dfrac{moles Pb(NO_{3})_{2}}{0.0745L}\]

\[(0.108M)(0.0745L)=0.00814molesPb(NO_{3})_{2}\]

Step 3: \[0.00814molsPb(NO_{3})_{2}\times \dfrac{2molsKI}{1molPb(NO_{3})_{2}}=0.0163molsKI\]

Step 4: \[0.155M=\dfrac{0.0163molsKI}{Liters}\]

\[\dfrac{0.0163molsKI}{0.155M}=0.105LitersKI\]

## 10.60

As a lab experiment the class submerged a 47.3-g aluminum block with an initial temperature of 32.4 degrees C into an unknown mass of water at 73.2 degrees C. The temperature of the final mixture at equilibrium is 57.3 degrees C. Find the unknown mass of the water used in the experiment.

__Definitions:__

Specific Heat (C_{s}): the required amount of heat energy to increase the temperature of one gram of a certain substance by 1 degree C.

__Helpful Equations:__

Heat gained by aluminum = heat lost by water

\[Heat(Q)=mass\times C_{s}\times \Delta T\]

*Hint: Do not convert temperature to Kelvin

*Hint: Change in Temperature = T_{final }- T_{initial}

Specific heat of water = 4.18 J/g x C

Specific heat of aluminum = 0.903 J/g x C

Units:

Mass = grams

Specific Heat = J/g x C

Temperature = Celsius

__Solving Strategy:__

Step 1: Due to the Law of Conservation of Energy, the heat exchange of aluminum and water can be set equal to each other. We know that aluminum will gain heat while water looses heat until they reach an equilibrium temperature.

Step 2: Set up the equation as:

\[mass_{water}\times C_{swater}\times \Delta T_{water}=-mass_{Al}\times C_{sAl}\times \Delta T_{Al}\]

Step 3: Now plug in known values and solve.

__Solution:__

Solve: \[mass_{water}\times 4.18J/g\cdot ^{\circ}C\times (53.7^{\circ}C-73.2^{\circ}C)=-47.3g\times (0.903J/g\cdot ^{\circ}C)\times (53.7^{\circ}C-32.4^{\circ}C)\]

\[Mass_{water}=\dfrac{-47.3g\times 0.903J/g\cdot ^{\circ}C\times (53.7^{\circ}C-32.4^{\circ}C)}{4.18J/g\cdot ^{\circ}C\times (53.7^{\circ}C-73.2^{\circ}C)}\]

\[mass_{water}=11.16g\]