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AVprozaik [17]
3 years ago
15

1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%

Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?
Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)
Chemistry
1 answer:
Tatiana [17]3 years ago
3 0

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

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Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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Answer:

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Alkali

Explanation:

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Sodium hydroxide or potassium hydroxide is generally used as an alkali. The use of alkali depends on the intended application of the soap.

Raw animal fat was used in the past but these days, processed fat is used in the soap manufacturing process. Vegetable fats ( e.g, palm oil, olive oil, coconut oil) are also being used in soap manufacturing.

Additives are also used to enrich the color and texture of the soap.

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