Chemistry, 25.07.2019 23:00 stopithelpmeplease

If 2.00g of p-aminophenol ( 109.1 g/mol) reacts with 5.00 ml of acetic anhydride (102.1 g/mol and density = 1.08 g/ml), what mass of acetaminophen (180.2 g/mol) would be made? show all calculations.

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The complete reaction is as,

4-Aminophenol + Acetic Anhydride → Acetaminophen + Acetic Acid

First of all convert the ml of Acetic anhydrite to grams,
As,
Density = mass / volume
Solving for mass,
mass = Density × Volume
Putting values,
mass = 1.08 g/ml × 5ml

mass = 5.4 g of acetic anhydride

First Find amount of acetic anhydride required to react completely with 2 g of p-Aminophenol,
As,
109.1 g of p-aminophenol required = 102.1 g of acetic anhydride
so, 2 g of p-aminophenol will require = X g of Acetic Anhydride

Solving for X,
X = (2 g × 102.1 g) ÷ 109.1 g

X = 1.87 g of acetic anhydride is required to be reacted.

But, we are provided with 5.4 g of Acetic Anhydride, means p-aminophenol is the limiting reactant and it controls the formation of product. Now Let's calculate for product,
As,
109.1 g of p-aminophenol produced = 180.2 g of Acetaminophen
So 2.00 g of p-aminophenol will produce = X g of Acetaminophen

Solving for X,
X = (2.00 g × 180.2 g) ÷ 109.1 g

X = 3.30 g of Acetaminophen

Result:
If 2.00g of p-aminophenol reacts with 5.00 ml of acetic anhydride 3.30 g of acetaminophen is made.

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

Qand k are only the same when the forward rate of the reaction has reached zero.

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If 2.00g of p-aminophenol ( 109.1 g/mol) reacts with 5.00 ml of acetic anhydride (102.1 g/mol and de...

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