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alexgriva [62]
3 years ago
15

A student adds 200.0g of C7H6O3 to an excess of C4H6O3, this produces C9H8O4 and C2H4O2. Calculate the percent yield if 231 g of

aspirin (C9H8O4) is produced in an experiment.
___C7H6O3 + ___C4H6O3  ___C9H8O4 + ___C2H4O2
Chemistry
1 answer:
a_sh-v [17]3 years ago
4 0

Answer:

Percent yield = 88%

Explanation:

Given data:

Mass of C₇H₆O₃ = 200.0 g

Actual yield of aspirin = 231 g

Percent yield  = ?

Solution:

Chemical equation:

C₇H₆O₃  + C₄H₆O₃      →       C₉H₈O₄ + C₂H₄O₂

Number of  moles of aspirin:

Number of moles = mass/molar mass

Number of moles = 200.0 g/ 138.12 g/mol

Number of moles =  1.45 mol

Now we will compare the moles of aspirin with C₇H₆O₃.

                          C₇H₆O₃           :           C₉H₈O₄

                               1                 :               1

                            1.45              :             1.45

Theoretical yield of aspirin:

Mass = number of moles × molar mass

Mass = 1.45 g × 180.158 g/mol

Mass = 261.23 g

Percent yield:

Percent yield =( actual yield / theoretical yield )× 100

Percent yield =  (231 g/ 261.23 g)× 100

Percent yield = 0.88 × 100

Percent yield = 88%

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SVEN [57.7K]

Answer:

The pH is equal to 4.41

Explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

                HClO   ⇄   ClO-  +  H+

start:          0.05            0         0

change       -x               +x       +x

balance     0.05-x         x         x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

3x10^{-8}=\frac{x^{2} }{0.05}

clearing the x and calculating its value we have:

x=3.87x10^{-5}=[H+]=[ClO-]

the pH can be calculated by:

pH=-log[H+]=-log[3.87x10^{-5}]=4.41

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Which of these is an example of a wetland area? A. marsh B. stream C. city park D. farm land
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At which concentration of glucose carriers was the glucose concentration reduced to zero?
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How many milliliters of 0. 0850 m naoh are required to titrate 25. 0 ml of 0. 0720 m hbr to the equivalence point?
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The volume of 0.0850 M NaOH is required to titrate 25.0 ml of 0.0720 M is 21.176 ml.

<h3>What are normality and molarity?</h3>

Normality is defined as the number of equivalents per litre of the solution

It is given by

Normality = number of equivalents / 1 L of solution

Molarity is defined as the number of moles of solute present in one litre of solution

Molarity = moles of solute / 1 litre of solution

Relationship between molarity and normality

Normality = Molarity × Acidity or Basicity of a salt

Here, the Acidity Of NaOH is 1 and the basicity of HBr is also 1.

Thus, Normality = Molarity

We know, N₁V₁ = N₂V₂

We can also write it as M₁V₁ = M₂V₂

V1 = \frac{M_2V_2}{M_!}

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V1 = 21.176 ml

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During your reaction, you added 0.3 mL concentrated H2SO4 (18.4 M) as the catalyst. At the end of the reaction, you need to add
sattari [20]

Answer:

58.72 mL

Explanation:

The chemical equation for the neutralization reaction is :

H₂SO₄(aq) + Na₂CO₃(s)  --------------> Na₂SO₄(aq) + H₂O(l) + CO₂(g)

where;

M₁ = Molarity of H₂SO₄

M₂= Molarity of Na₂CO₃

V₁= Volume of H₂SO₄

V₂ = Volume of Na₂CO₃

Given that :

M₁ = 18.4 M

V₁= 0.3 mL

10% Na₂CO₃ means 100 g of solution contain 10 g of Na₂CO₃

i.e. 10 g Na₂CO₃ dissolved and diluted to 100 mL water.

Molar mass of Na₂CO₃ = 106 g/mol

106 g Na₂CO₃ dissolved in 100 mL will give 0.1 M Na₂CO₃ solution.

However;

If, 106 g Na₂CO₃ ≡ 0.1 M Na₂CO₃

Then, 10 g Na₂CO₃ ≡  'A' M  of Na₂CO₃

By cross multiplying; we have:

106 × A = 10 × 0.1

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A = (1/106) M/100 mL

A = 10 x (1/106)) M/L

A = (10/106) M

A = 0.094  M

Therefore,the molarity of 10% Na₂CO₃ solution is 0.094 M.

For the Neutralization equation, we have:

M₁V₁ = M₂V₂

18.4×0.3 = 0.094×V₂

Making V₂  the subject of the formula;we have:

V_2 = \frac{18.4*0.3}{0.094}

V₂ = 58.72 mL

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3 years ago
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