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3 years ago

What is Stoichiometric Calculations

Chemistry

2 answers:

ANTONII [103]3 years ago

4 0

Answer:

Stoichiometry is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. I think.

Sophie [7]3 years ago

4 0

Answer: Stoichiometry is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation.

Explanation:

Stoichiometric Calculations are mostly based on chemical formulas. Formula Mass: It is defined as the sum of the atomic weights of each atom present in the molecule of the substance. For example formula mass of Na2S is calculated as 2(23) + 1(32) = 78.

Hope it help <3

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4.94x10^24 atoms Zn

Explanation:

By definition, there are 6.022x10^23 atoms in a mole of a substance. This is a conversion factor:

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4 years ago

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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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