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3 years ago

Calculate the mass of ethyl alcohol required to prepare 540 grams of C4H6, if the reaction follows the scheme:

Chemistry

1 answer:

kkurt [141]3 years ago

3 0

Answer:

Explanation:

2C₂H₅OH = C₄H₆ + 2H₂O + H₂

2 mole 1 mole

molecular weight of ethyl alcohol

mol weight of C₂H₅OH = 46 gm

mol weight of C₄H₆ 54 gm

540 gm of C₄H₆ = 10 mole

10 mole of C₄H₆ will require 20 mol of ethyl alcohol .

20 mole of ethyl alcohol = 20 x 46

= 920 gm

ethyl alcohol required = 920 gm .

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A piece of metal with a mass of 159 g is placed in a 50 mL graduated cylinder. The water level rises from 20. mL to 41 mL. What

almond37 [142]

Volume of water displaced is equal to the volume of the metal. So 41 - 20 = 21 ml. Density = mass / volume so density = 159 / 21 = 7.57 g/ml

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3 years ago

True or false certain cations are associated with either exotermic or endothermic processes

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Explanation:

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3 years ago

Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3

Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

$Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }$

You know:

Kc= 14.5
[H₂]= 0.322 M
[CH₃OH] =1.56 M

Replacing:

$14.5=\frac{1.56}{[CO]*0.322^{2} }$

Solving:

$[CO]=\frac{1.56}{14.5*0.322^{2} }$

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

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3 years ago

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vladimir2022 [97]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

In which of the following reactions does a decrease in the volume of the reaction vessel at constant

Black_prince [1.1K]

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

D) 2O₃(g) → 3O₂(g)

The number of moles of reactant is 2 and number of moles of product is 3.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

6 0

3 years ago