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2 years ago

Calculate the mass of oxygen produced from the decomposition of 75.0 g of potassium chlorate?

Chemistry

2 answers:

Dmitry [639]2 years ago

4 0

The balanced reaction is:

2KClO3 = 2KCl + 3O2

We are given the amount of potassium chlorate. This will be our starting point.

75.0 g KClO3 (1 mol KClO3 /122.55 g KClO3) (3 mol O2/2 mol KClO3) ( 32 g O2 / 1mol O2) = 58.75 g O2

Thus, the answer is 28.88g H2O.

Elis [28]2 years ago

4 0

Answer:

29.4 g

Explanation:

Let's consider the decomposition of potassium chlorate.

KClO₃ → KCl + 1.5 O₂

The molar mass of KClO₃ is 122.55 g/mol. The moles of KClO₃ in 75.0 g are:

75.0 g × (1 mol/ 122.55 g) = 0.612 mol

The molar ratio of KClO₃ to O₂ is 1:1.5. The moles of O₂ are 1.5 × 0.612 mol = 0.918 mol

The molar mass of O₂ is 32.00 g/mol. The mass of O₂ corresponding to 0.918 moles is:

0.918 mol × (32.00 g/mol) = 29.4 g

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Answer:

One

Explanation:

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2 years ago

"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

Salsk061 [2.6K]

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HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

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E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$ [/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

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Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

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So, the value of Ka for butyric acid is $1.53*10^-^5$ .

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