Question

I NEED HELP, THANKS!

Using the Ideal Gas Law, PV = nRT, where R = 0.0821 L atm/mol K, calculate the volume in liters of oxygen produced by the catalytic decomposition of 25.5 g potassium chlorate according to the following reaction. The oxygen is collected at 2.22 atm and 25.44°C. Express your answer to the correct number of significant figures.

Answer 1

Solution:-

The balanced equation:

2KClO3 (s) \rightarrow 2KCl (s) + 3 O2 (g)

Molar mass of KClO3 = 122.55 g/mol

Number of moles of KClO3 = (Mass of KClO3 / Molar mass of KClO3) = (25.5 g / 122.55 g/mol) = 0.2081 mol

From balanced equation, 2 mol of KClO3 produce 3 mol of O2. Or, 1 mol of KClO3 produces (3/2) mol of O2.

therefore, 0.2081 mol of KClO3 will produce = (3/2) × (0.2081) = 0.3121 mol of O2

Now, we have number of moles (n) of O2 = 0.3121 mol

Pressure (P) = 2.22 atm

Temperature (T) = 25.44°C = (273.15 + 25.44) K = 298.59 K

R (Gas constant) = 0.0821 L.atm/mol.K

Volume (V) of O2 = ?

Using the ideal gas equation,

V = nRT / P

Substituting the values in the equation, we get :

V = (0.3121 mol × 0.0821 L.atm/mol.K × 298.59 K) / (2.22 atm) = 3.45 L

Hence, the volume of O2 gas produced = 3.45 L

Step-by-step explanation:

Answer 2

`\large \boxed{\text{3.45 L}}`

Step-by-step explanation:

We will need a balanced chemical equation with molar masses, so, let's gather all the information in one place.

Mᵣ: 122.55

2KClO₃ ⟶ 2KCl + 3O₂

m/g: 25.5

(a) Moles of KClO₃

`\text{Moles of KClO}_(3) =\text{25.5 g KClO}_(3) * \frac{\text{1 mol KClO}_(3)}{\text{122.55 g KClO}_(3)} = \text{0.2081 mol KClO}_(3)`

(b) Moles of O₂

The molar ratio is 3 mol O₂:2 mol KClO₃

`\text{Moles of O$_(2)$}= \text{0.2081 mol KClO}_(3) * \frac{\text{3 mol O$_(2)$}}{ \text{2 mol KClO}_(3)} = \text{0.3121 mol O$_(2)$}`

(c) Volume of O₂

We can use the Ideal Gas Law to calculate the volume of hydrogen.

pV = nRT

T = (25.44 + 273.15) K = 298.59 K

`\rm V = (nRT)/(p)= \frac{\text{0.3121 mol $*$ 0.0821 L$\cdot$atm$\cdot$K$^(-1)$mol$^(-1)*$ 298.59 K}}{\text{ 2.22 atm}} = \textbf{3.45 L} \\\\\text{The volume of oxygen is $\large \boxed{\textbf{3.45 L}}$}`