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GenaCL600 [577]
3 years ago
5

How many grams of sodium oxide can be synthesized from 17.4 g of sodium? assume that more than enough oxygen is present. the bal

anced equation is: 4na(s) + o2(g) → 2na2o(s)?
Chemistry
1 answer:
Tatiana [17]3 years ago
7 0
Equation is as follow,

<span>                          4 Na (s)  +  O</span>₂ <span>(g)    →    2Na</span>₂<span>O (s)

According to equation,

      91.92 g (4 moles) of Na produces  =  123.92 g (2 moles) of Na</span>₂O
So,
                    17.4 g of Na will produce  =  X g of Na₂O

Solving for X,
                                   X  =  (17.4 g × 123.92 g) ÷ 91.92 g

                                   X  =  23.45 g of Na₂O
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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
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This is an incomplete question, here is a complete question.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)

Substance      ΔG°f (kJ/mol)

M₃O₄                 -9.50

M(s)                       0

O₂(g)                     0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

PCl_3(l)\rightarrow PCl_3(g)

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

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]

where,

\Delta G^o = Gibbs energy of reaction = ?

n = number of moles

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

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]

\Delta G^o=9.50kJ/mol

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs free energy  = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K  = equilibrium constant = ?

9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)

K=0.0216

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

K=(P_{O_2})^2

0.0216=(P_{O_2})^2

P_{O_2}=0.147atm

The equilibrium pressure of O₂(g) is, 0.147 atm

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