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Mice21 [21]
3 years ago
15

Hydrogen gas and gaseous iodine will react to form hydrogen iodide, as described by the following chemical equation.H2(g)+I2(g)↽

−−⇀HI(g)Kc(400∘C)=50.0Assume that all of the HI(g) is removed from a vessel containing this reaction, and equilibrium is re-established. What will be the new equilibrium concentration of HI if the equilibrium concentrations of H2 and I2 are both equal to 0.450M?
Chemistry
1 answer:
Kisachek [45]3 years ago
4 0

Answer:

3.18 M

Explanation:

Given data

[H₂]eq = [I₂]eq = 0.450 M

[HI]eq = ?

Let's consider the following reaction at equilibrium.

H₂(g) + I₂(g) ⇄ 2 HI(g)      Kc(400°C) = 50.0

The concentration equilibrium constant (Kc) is:

Kc = [HI]²eq / [H₂]eq × [I₂]eq

[HI]²eq = Kc × [H₂]eq × [I₂]eq

[HI]eq = √(Kc × [H₂]eq × [I₂]eq)

[HI]eq = √(50.0 × 0.450 × 0.450)

[HI]eq = 3.18 M

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1) divide each percentage by the relative atomic mass of the element

2) divide all results by the smallest number

3)multiply by a whole number to get the simplest whole number ratio (if necessary)

that is to say:

Na                                   S                            O

32.37÷23                 22.58÷32                45.05÷16

= 1.407                       = 0.7056               = 2.816       (to 4 significant figures)

the smallest number here is 0.7056 so:

1.407÷0.7056        0.7056÷0.7056       2.816÷0.7056

=1.99 approx.2                   = 1                   3.99 approx. 4

here there is no need to carry out step 3 as ratio obtained is already a simplest whole number  ratio

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8 0
3 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
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Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

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Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

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