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blsea [12.9K]
4 years ago
10

How many grams of chlorine gas can be produced if 15 grams of FeCl3 reacts with 4 moles of O2? What is the limiting reactant? Wh

at is the excess reactant? If 9.5 grams of Cl2 gas is produced, what is the percent yield?
4 FeCl3 + 3O2 --> 2Fe2O3 + 6Cl2
In paragraph form write out the steps used to solve the above stoichiometry calculation. Explain how do we find the limiting reactant, excess reactant, theoretical yield, actual yield and calculate the percent yield?
Chemistry
1 answer:
Alika [10]4 years ago
6 0

Answer:

dang this goformative assignment got you too huh.. whatever happened to "Easy writing assignment" .... this is a whole test in itself. SMH!

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Answer: eukaryotic cell

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3 years ago
H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles
abruzzese [7]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

H_2(g)+F_2(g)\rightarrow 2HF(g)

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]

We are given:

\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = \frac{-14.1}{1}\times 2.20=-31.02J/K

Hence, the value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

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Answer:

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Explanation:

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8 0
3 years ago
Butane combusts in the atmosphere and releases heat:2C4H10(g) + 13 O2(g) ---&gt; 8 CO2(g) + 10 H2O(g)The signs of the values for
anastassius [24]

Answer:\Delta G: -ve

\Delta H : -ve,

\Delta S : +ve

Explanation:

Endothermic reactions are those in which heat is absorbed by the system and exothermic reactions are those in which heat is released by the system.

\Delta H for Endothermic reaction is positive and  \Delta H for Exothermic reaction is negative.

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

As 15 moles of gaseous reactants are changing into 18 moles of gaseous products, randomness is increasing and thus \Delta S is positive.

Using Gibbs Helmholtz equation:

\Delta G=\Delta H-T\Delta S

\Delta G=(-ve)-T(+ve)

\Delta G=(-ve)(-ve)=-ve

Thus \Delta H is negative , \Delta S is positive and \Delta G is negative.

5 0
3 years ago
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