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stira [4]
3 years ago
5

What is the mass of 5.26 mol Fe2(SO4)3? Answer in units of g.

Chemistry
1 answer:
dsp733 years ago
3 0
<h3>Answer:</h3>

2100 g Fe₂(SO₄)₃

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis

<u>Atomic Structure</u>

  • Reading a Periodic Table
<h3>Explanation:</h3>

<u>Step 1: Define</u>

5.26 mol Fe₂(SO₄)₃

<u>Step 2: Identify Conversions</u>

Molar Mass of Fe - 55.85 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe₂(SO₄)₃ - 2(55.85) + 3(32.07) + 12(16.00) = 399.91 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 5.26 \ mol \ Fe_2(SO_4)_3(\frac{399.91 \ g \ Fe_2(SO_4)_3}{1 \ mol \ Fe_2(SO_4)_3})
  2. Multiply/Divide:                 \displaystyle 2103.53 \ g \ Fe_2(SO_4)_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

2103.53 g Fe₂(SO₄)₃ ≈ 2100 g Fe₂(SO₄)₃

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(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

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Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

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H

3

O

+

m

+

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:

m

m

0.05

m

m

m

m

m

m

m

m

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0

m

m

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m

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l

l

0

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:

m

m

l

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m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

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:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

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[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

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l

mol/L

pH

=

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[

H

3

O

+

]

=

-log

(

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×

10

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)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

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[

A

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.

pH

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p

K

a

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log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

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10

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)

+

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(

[

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0.05

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=

3.49

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log

(

[

A

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0.05

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log

(

[

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0.05

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0.05

=

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=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

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2 years ago
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Answer:

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