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matrenka [14]
2 years ago
11

4 b + 3 o2 → 2 b2o3 if 8 moles of b and 4 moles of o2 are allowed to react, how many moles of b2o3 can be formed?

Chemistry
1 answer:
raketka [301]2 years ago
6 0

The number of moles of  b2o3 that will be formed is determined as 4 moles.

<h3>Limiting reagent</h3>

The limiting reagent is the reactant that will be completely used up.

4 b + 3O₂ → 2b₂O₃

from the equation above;

4 b ------------> 2 b₂O₃

2b ------------> b₂O₃

2 : 1

3O₂  -------------> 2b₂O₃

3  :  2

b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;

4 b ------------> 2 moles of  b2o3

8 moles -------> ?

= (8 x 2)/4

= 4 moles

Thus, the number of moles of  b2o3 that will be formed is determined as 4 moles.

Learn more about limiting reactants here: brainly.com/question/14222359

#SPJ1      

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7 0
2 years ago
Use Hess's Law to determine the enthalpy change (∆H) for the reaction: ClF + F2 → ClF3 Given: 2ClF + O2 → Cl2O + F2O. ∆H=167.4kJ
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The enthalpy change (∆H) for the reaction is -108.7 kJ

Explanation:

Hess's law can be stated as: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps. Then, Hess's Law states that the enthalpy of one reaction can be achieved by algebraically adding the enthalpies of other reactions.

So,  to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them.

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

2 ClF + O₂ → Cl₂O + F₂O ∆H=167.4kJ

Cl₂O + 3 F₂O → 2 ClF₃ + 2 O₂ ∆H= -341.4kJ  

The previous equation must be inverted, and the enthalpy value is also inverted, that is, the sign is changed.

2 F₂ + O₂ →2 F₂O ∆H=-43.4kJ

Reactants and products are added or canceled, taking into account that certain substances sometimes appear as a reagent and others as a product, so they are totally eliminated (there is nothing left of them anywhere in the reaction, if the same amount in reagents and products) or partially (this substance remains, in less quantity, only on one side), obtaining:

2 ClF + 2 F₂ → 2 ClF₃

Then, as all the reactants and products have a stoichiometric coefficient of 2, dividend by that number is obtained:

ClF + F₂ → ClF₃

Adding the enthalpies algebraically, and dividing by 2, because to get the "data" reaction you had to multiply by two, you get:

ΔH= [167.4 kJ - 341.4 kJ - 43.3 kJ]÷2

ΔH= -108.7 kJ

<u><em>The enthalpy change (∆H) for the reaction is -108.7 kJ</em></u>

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3 years ago
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