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4 years ago

Explain how you would make 450 mL of a .250M NaOH solution?

Chemistry

1 answer:

Neporo4naja [7]4 years ago

4 0

Answer:

Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.

Explanation:

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of NaOH = x

Molar mass of NaOH = 40 g/mol

Volume of the NaOH solution = 450 mL =- 0.450 L ( 1 ml = 0.450 L)

Molarity of the solution of NaOH = 0.250 M

$Molarity=\frac{1.248 g}{26 g/mol\times 0.9102254 L}=0.528 mol/L$

$0.250 M=\frac{x}{40 g/mol\times 0.450 L}$

Solving for x:

x = 4.5 g

Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.

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Given the following equation, what would be the theoretical yield in grams of acetylsalicylic acid (C9H8O4) that can be produced

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Answer:

The correct answer is 2.6 g C₉H₈O₄

Explanation:

-First we have to write and to balance the chemical equation:

C₇H₆O₃ + C₄H₆O₃ ⇄ C₉H₈O₄ + C₂H₄O₂

-Then, we calculate the molecular masses of reactants and products:

1 mol C₇H₆O₃= (12 g/mol C x 7) + (1 g/mol H x 6) + (16 g/mol O x 3)= 138 g

1 mol C₄H₆O₃= (12 g/mol C x 4) + (1 g/mol H x 6) + (16 g/mol O x 3)= 102 g

1 mol C₉H₈O₄= (12 g/mol C x 9) + (1 g/mol H x 8) + (16 g/mol O x 4)= 180 g

1 mol C₂H₄O₂= (12 g/mol C x 2) + (1 g/mol H x 4) + (16 g/mol O x 2)= 60 g

The mass balance is correct because:

mass reactants = mass products

138 g + 102 g = 180 g + 60 g

240 g = 240 g

-Now we use the masses from the chemical equation to calculate how reactant we need. We know that 138 g of salicylic acid (C₇H₆O₃) react with 102 g of acetic anhydride (C₄H₆O₃). So, the grams of acetic anhydride we need to react with 2 g of salicylic acid will be:

138 g C₇H₆O₃------------------- 102 g C₄H₆O₃

2.0 g C₇H₆O₃ -------------------- x= (2.0 x 102)/138 = 1.48 g C₄H₆O₃

If we compare, the amount of C₄H₆O₃ we need (1.48 g) is lesser than the amount we have (8 g), so C₄H₆O₃ is the excess reactant and C₇H₆O₃ is the limiting reactant.

- Finally, we use the limiting reactant to calculate the theoretical yield in grams of C₉H₈O₄. From the chemical equation, we know that 138 g C₇H₆O₃ yield 180 g of C₉H₈O₄. We have 2.0 g, so:

138 g C₇H₆O₃------------------- 180 g C₉H₈O₄

2.0 g C₇H₆O₃ -------------------- x= (2.0 x 180)/138 = 2.6 g C₉H₈O₄

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Answer is 355 grams.
Explanation:
Given the molecular weights:
M
r
N
a
O
H
=
40
g
m
o
l

M
r
N
a
2
S
O
4
=
142
g
m
o
l
The analogy of the moles will be held constant:
n
N
a
O
H
n
N
a
2
S
O
4
=
2
1
n
N
a
O
H
n
N
a
2
S
O
4
=
2
For each one, substitute:
n
=
m
M
r
Therefore:
n
N
a
O
H
n
N
a
2
S
O
4
=
2
m
N
a
O
H
M
r
N
a
O
H
m
N
a
2
S
O
4
M
r
N
a
2
S
O
4
=
2
200
40
x
142
=
2
200
⋅
142
40
x
=
2
200
⋅
142
=
2
⋅
40
x
x
=
200
⋅
142
2
⋅
40
=
100
⋅
142
40
=
10
⋅
142
4
=
1420
4
=
=
710
2
=
355
g
r
a
m
s
(or just use a calculator)

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