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pochemuha
3 years ago
5

I need help on 4, this is theoretical yeild and writing the equation

Chemistry
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

2.26 g

Explanation:

Data given:

mass of Sulfuric acid = 3.24 g

mass of Aluminum hydroxide = 0.945 g

Theoretical  yield of Aluminum sulfate = ?

Solution:

First we look for the balance reaction

Reaction

     3H₂SO₄ +  2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O

Now we look for the limiting reactant on which the amount of aluminum sulfate depends

So,

       3H₂SO₄ +  2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O

       3 mol          2 mol

from above reaction it is clear that 3 mole of H₂SO₄ combine with 2 mole of Al(OH)₃

Convert moles to mass

  • molar mass of Al(OH)₃  

molar mass of Al(OH)₃  =  27 + 3(16 + 1)

molar mass of Al(OH)₃  = 27 + 45

molar mass of Al(OH)₃  =72 g/mol

  • molar mass of H₂SO₄

molar mass of H₂SO₄ = 2(1) + 32 + 4(16)

molar mass of H₂SO₄ = 2 + 32 + 64

molar mass of H₂SO₄ = 98 g/mol

So,

      3H₂SO₄         +      2Al(OH)₃     ---------->   Al₂(SO₄)₃ + 6H₂O

   3 mol (98 g/mol)    2 mol (72 g/mol)

        294 g                      144 g

So its clear from the reaction that  294 g of H₂SO₄ react with 144 g of Al(OH)₃

now if we look at the given amounts the amount Al(OH)₃ is less then the amount of H₂SO₄

So, for Al(OH)₃ if we calculate the needed amount of H₂SO₄

So apply unity formula

            294 g H₂SO₄ ≅ 144 g of Al(OH)₃

             X g H₂SO₄ ≅ 0.945 of Al(OH)₃

Do cross multiplication

            X g H₂SO₄ = 294 g x 0.945 g / 144 g

            X g of H₂SO₄ ≅ 1.93 g

So, 1.93 g of H₂SO₄ will react out of 3.24 grams, the remaining amount of it will be in excess.

So,

Al(OH)₃ will be consumed completely an it will be limiting reactant.

-----------

Now to Calculate for the theoretical yield

First we look for the balance reaction

Reaction

     3H₂SO₄ +  2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O

Now we look for the mole mole ration of Al(OH)₃ to the amount of aluminum sulfate produced

So,

       3H₂SO₄ +  2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O

                          2 mol                       1 mole

from above reaction it is clear that 1 mole of Al₂(SO₄)₃ produce by 2 mole of Al(OH)₃

As we know that

2 mole of Al(OH)₃ = 144 g

So,

if 144 g of Al(OH)₃ gives 1 mole Al₂(SO₄)₃  then how many moles of Al₂(SO₄)₃  will be produces by 0.945 g Al(OH)₃

So apply unity formula

            144 g of Al(OH)₃ ≅ 1 mole of Al₂(SO₄)₃

            0.945 g of Al(OH)₃ ≅ X mole of Al₂(SO₄)₃

Do cross multiplication

           X mole of Al₂(SO₄)₃ = 0.945 g  x 1 mole / 144 g

            X mole of Al₂(SO₄)₃ = 0.0066 moles

So,

0.945 g of  Al(OH)₃ produce 0.0066 mole of Al₂(SO₄)₃

Now conver moles of Al₂(SO₄)₃ to mass

Formula used:

       mass in grams = no. of moles x molar mass . . . . . . (1)

  • molar mass of Al₂(SO₄)₃  

molar mass of Al₂(SO₄)₃ =  2(27) + 3(32 +4(16))

molar mass of Al₂(SO₄)₃ = 54 + 3 (32 +64)

molar mass of Al₂(SO₄)₃ = 54 + 3 (96)

molar mass of Al₂(SO₄)₃ = 54 + 288

molar mass of Al₂(SO₄)₃  =342 g/mol

Put values in equation 1

        mass in grams = 0.0066 g x 342 g/mol

        mass in grams = 2.26 g

So the theoretical yield of Aluminum sulfate (Al₂(SO₄)₃ ) is 2.26 g

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Answer:

158 L.

Explanation:

What is given?

Pressure (P) = 1 atm.

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Molar mass of methane CH4 = 16 g/mol.

R constant = 0.0821 L*atm/mol*K.

What do we need? Volume (V).

Step-by-step solution:

To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:

PV=nRT.

Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.

So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

80.0g\text{ CH}_4\cdot\frac{1\text{ mol CH}_4}{16\text{ g CH}_4}=5\text{ moles CH}_4.

So, in this case, n=5.

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V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}

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2. Extensive properties:

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Specific heat is the amount of heat required to increase the temperature of any substance per unit mass. Specific heat capacity is also known as mass specific heat. Its SI unit is Joule (J).

The formula to calculate the heat energy of copper is as follows:

                                       …… (1)

Here,

Q is the amount of heat transferred.

m is the mass of copper.

c is the specific heat of copper.

is the change in temperature of copper.

Rearrange equation (1) to calculate the temperature change.

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The value of Q needs to be converted into J. The conversion factor for this is,

So the value of Q can b calculated as follows:

The value of Q is 4689 J.

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The value of c is .

Substitute these values in equation (2).

The temperature change  can be calculated as follows:

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Here,

is the change in temperature.

is the final temperature.

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Rearrange equation (3) to calculate the final temperature.

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So the final temperature of copper is .

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The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate
Nimfa-mama [501]

Answer:

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atomic radius = 2.3103x10^-8 cm

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p=0.855 g/cm^3

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r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3}  }{4}=2.3103x10^{-8}cm

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3 years ago
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