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3 years ago

Calculate the moles of SO₂ produced when 5 moles of FeS₂ reacts according to the equation: 4FeS₂+ 11O₂ → 2Fe₂O₃ + 8SO₂

Chemistry

1 answer:

Gala2k [10]3 years ago

7 0

Answer:

10 moles of SO₂ are produced when 5 moles of FeS₂

Explanation:

Stoichiometry: it is the theoretical proportion in which the chemical species are combined in a chemical reaction. The stoichiometric equation of a chemical reaction relates molecules or number of moles of all the reagents and products that participate in the reaction.

In other words, stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships established are molar relationships (that is, moles) between the compounds or elements that make up the chemical equation.

The stoichiometric coefficients of a chemical reaction indicate the proportion in which said substances react.

Taking into account the above, you can apply the following rule of three: by stoichiometry if 4 moles of FeS₂ produce 8 moles of SO₂, then when reacting 5 moles of FeS₂ how many moles of SO₂ will they produce?

$moles of SO_{2} =\frac{5 moles of FeS_{2} *8moles of SO_{2}}{4 moles of FeS_{2}}$

moles of SO₂= 10

10 moles of SO₂ are produced when 5 moles of FeS₂

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Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

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AlexFokin [52]

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Determine the expression for the equilibrium constant, KcKcK_c, for the reaction by identifying which terms will be in the numer

Yakvenalex [24]

Answer: Hello your question lacks some details below is the complete question

answer :

Numerator = CH₃OH

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hence for the equilibrium constant Kc

Numerator = CH₃OH

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Elena-2011 [213]

Answer:

0.0646 mole.

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The following data were obtained from the question:

Number of molecules = 3.89×10²² molecules

Number of mole =?

The number of mole present in 3.89×10²² molecules of CCl₄ can be obtained as follow:

From Avogadro's hypothesis:

6.02×10²³ molecules = 1 mole

Therefore,

3.89×10²² molecules = 3.89×10²² molecules × 1 mole / 6.02×10²³ molecules

3.89×10²² molecules = 0.0646 mole

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