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Zina [86]
3 years ago
11

Consider the reaction: 2 SO2(g) + O2(g) <----> 2 SO3. If, at equilibrium at a certain temperature, [SO2] = 1.50 M, [O2] =

0.120 M, and [SO3] = 1.25 M, what is the value of the equilibrium constant? none of the above 8.68 6.94 0.14 5.79
Chemistry
1 answer:
vlada-n [284]3 years ago
7 0

Answer:

5.79

Explanation:

The following data were obtained from the question:

Concentration of SO2, [SO2] = 1.50 M

Concentration of O2, [O2] = 0.120 M

Concentration of SO3, [SO3] = 1.25 M

Equilibrium constant, (K) =..?

The balanced equation for the reaction is given below:

2SO2(g) + O2(g) <==> 2SO3(g)

From the balanced equation, the equilibrium constant K is written as follow:

K = [SO3]²/ [SO2]²• [O2]

With the above, the value of the equilibrium constant, K can be obtained as follow

K = [SO3]²/ [SO2]²• [O2]

K = (1.25)² / (1.50)² × 0.120

K = 1.5625 / ( 2.25 x 0.120)

K = 5.79

Therefore, the equilibrium constant is 5.79.

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Consider the balanced chemical equation for the combustion of methane (CH4).
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<h3>Answer:</h3>

                  89.6 L of O₂

<h3>Solution:</h3>

The balanced chemical equation is as,

                                CH₄  +  2 O₂    →    CO₂  +  2 H₂O

As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

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If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
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Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

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From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

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We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

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