Respuesta :

Answer:

0.058824 g

Explanation:

To calculate the moles of oxygen gas formed, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 670 mmHg  

V = Volume of the gas = 20.0 mL = 0.02 L ( 1 mL= 0.001 L)

T = Temperature of the gas =293 K

R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

[tex]670mmHg\times 0.02L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{670\times 0.02}{62.3637\times 293}=0.00073mol[/tex]

According the reaction shown below:-

[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]

3 moles of oxygen gas is produced when 2 moles of potassium chlorate reacts

So,

1 mole of oxygen gas is produced when [tex]\frac{2}{3}[/tex] moles of potassium chlorate reacts

Also,

0.00073 mole of oxygen gas is produced when [tex]\frac{2}{3}\times 0.00073[/tex] moles of potassium chlorate reacts

Moles of potassium chlorate reacts = 0.00048 Moles

Molar mass of potassium chlorate = 122.55 g/mol

Mass = Moles*Molar mass = [tex]0.00048\times 122.55\ g[/tex] = 0.058824 g

Eduard22sly

The mass of potassium chlorate, KClO₃ needed to produce 20.0 mL

of oxygen gas at 670 mmHg and 293 K is 0.06 g

We'll begin by calculating the number of mole of O₂ produced from the reaction. This can be obtained as follow:

Volume (V) = 20 mL = 20 / 1000 = 0.02 L

Pressure (P) = 670 mmHg = 670 / 760 = 0.882 atm

Temperature (T) = 293 K

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

0.882 × 0.02 = n × 0.0821 × 293

0.01764 = n × 24.0553

Divide both side by side 24.0553

n = 0.01764 / 24.0553

n = 0.00073 mole

  • Next, we shall determine the number of mole of KClO₃ needed to produce 0.00073 mole of O₂

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂

Therefore,

X mole of KClO₃ will decompose to 0.00073 mole of O₂ i.e

X mole of KClO₃ = [tex]\frac{2 * 0.00073}{3}\\\\[/tex]

X mole of KClO₃ = 0.00049 mole

  • Finally, we shall determine the mass of 0.00049 mole of KClO₃.

Mole of KClO₃ = 0.00049 mole

Molar mass of KClO₃ = 39 + 35.5 + (3×16) = 122.5 g/mol

Mass of KClO₃ =?

Mass = mole × molar mass

Mass of KClO₃ = 0.00049 × 122.5

Mass of KClO₃ = of 0.06 g

Therefore, the mass of KClO₃ needed for the reaction is 0.06 g

Learn more: https://brainly.com/question/14790287

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