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givi [52]
3 years ago
6

PLEASE HELPP!!

Chemistry
1 answer:
Ivanshal [37]3 years ago
5 0

Answer:

Answer: The correct answer is Option B.

Explanation:

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

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

For N_2N2 :

Given mass of nitrogen gas = 10 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

\text{Moles of iron oxide}=\frac{10g}{28g/mol}=0.357molMoles of iron oxide=28g/mol10g=0.357mol

The given chemical reaction follows:

N_2+O_2\rightarrow 2NON2+O2→2NO

As, oxygen gas is present in excess. Thus, it is considered as an excess reagent and nitrogen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

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At a certain temperature, 0.800 mol SO 3 is placed in a 1.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil
Elanso [62]

Answer : The value of equilibrium constant (K) is, 0.004

Explanation :

First we have to calculate the concentration of SO_3\text{ and }O_2

\text{Concentration of }SO_3=\frac{\text{Moles of }SO_3}{\text{Volume of solution}}=\frac{0.800mol}{1.50L}=1.2M

and,

\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{0.150mol}{1.50L}=0.1M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                       2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

Initial conc.       1.2             0               0

At eqm.          (1.2-2x)        2x               x

As we are given:

Concentration of O_2 at equilibrium = x = 0.1 M

The expression for equilibrium constant is:

K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}

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

K_c=\frac{(2x)^2\times (x)}{(1.2-2x)^2}

K_c=\frac{(2\times 0.1)^2\times (0.1)}{(1.2-2\times 0.1)^2}

K_c=0.004

Thus, the value of equilibrium constant (K) is, 0.004

3 0
4 years ago
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