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andreev551 [17]
3 years ago
11

Balance the following reaction. A coefficient of \"1\" is understood. Choose option \"blank\" for the correct answer if the coef

ficient is \"1.\". . Al + ZnCl2 → Zn + AlCl3. . . 2.Balance the following reaction. A coefficient of \"1\" is understood. Choose option \"blank\" for the correct answer if the coefficient is \"1.\". . NH3 + O2 → N2 + H2O. .
Chemistry
1 answer:
g100num [7]3 years ago
8 0
Conservation of mass is the underlying principle of balancing equation. When we balance equation, this means that we acknowledge that before and after the chemical reaction, the elements are conserved. To balance the chemical equation, we add coefficients before each reactant and product. Here are the following answers: 

Reaction 1: 
<span>2Al
3ZnCl2
3Zn
2AlCl3
Reaction 2: 
</span><span>4NH3
3O2
2N2
6H2O</span>
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In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y 4 − Y4− ) and metal chelate (abbreviated MY n − 4 M
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There will 3.95 grams of Na2 and H2O that should be added to form a concentric required solution.

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Use the given data at 500 K to calculate ΔG°for the reaction
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Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(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_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

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

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

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_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

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

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

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 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

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3 0
3 years ago
What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?
AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

<u>Answer:</u> The theoretical yield of copper is 44.48 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of aluminium = 12.6 g

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Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol

The given chemical equation follows:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = \frac{3}{2}\times 0.467=0.7005mol of copper

Now, calculating the mass of copper  from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g

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