A student thermally decomposed a .810 gram sample of impure potassium chlorate. Manganese dioxide was used as a catalyst in the reaction. The student collected 43.60 mL of oxygen gas over water in a eudiometer. Potassium chloride was the other product of the reaction. The temperature and pressure at collection time were 20 Celcius and 762.10 torr respectively. The water level in the eudiometer was 4.22 cm above the outside water level in the beaker. a) write a balanced equation for the reaction. b) what is the corrected pressure of the dry oxygen gas c) what is the volume in mL of the dry oxygen gas at STP conditions d) how many molecules of oxygen gas were collected e) what is the percent purity of the original potassium chlorate sample.

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29-03-2021
Chemistry

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A student thermally decomposed a .810 gram sample of impure potassium chlorate. Manganese dioxide was used as a catalyst in the reaction. The student collected 43.60 mL of oxygen gas over water in a eudiometer. Potassium chloride was the other product of the reaction. The temperature and pressure at collection time were 20 Celcius and 762.10 torr respectively. The water level in the eudiometer was 4.22 cm above the outside water level in the beaker. a) write a balanced equation for the reaction. b) what is the corrected pressure of the dry oxygen gas c) what is the volume in mL of the dry oxygen gas at STP conditions d) how many molecules of oxygen gas were collected e) what is the percent purity of the original potassium chlorate sample.

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mickymike92 mickymike92 · Answer 1 · 2021-04-01T07:18:14+02:00

Answer:

a. 2 KClO3 ------> 2 KCl + 3 O2

b. Vapor pressure of dry oxygen gas = 741.45 torr

c. Volume at STP = 39.63 mL

d. Number of oxygen gas molecules = 1.065 × 10^21 molecules

e. Percent purity of KClO3 = 40.07 %

Explanation:

a) The balanced equation for the reaction is given below :

2 KClO3 ------> 2 KCl + 3 O2

b) Vapor pressure of dry oxygen gas = Total pressure - (water vapor pressure + pressure due difference in water levels)

Vapor pressure of water at 20 °C is 17.535 mm (torr).

Pressure due to difference in water level = 4.22 cm × 10mm/cm / 13.534 (13.534 is the density of mercury) = 3.118 mm (torr).

Vapor pressure of dry oxygen gas = 762.10 torr - 3.118 torr - 17.535 torr

Vapor pressure of dry oxygen gas = 741.45 torr

c) P1 = 741.45 torr

V1 = 43.60 ml

T1 = 20 °C + 273.15 = 293.15 K

P2 = 760 torr

T2 = 273.15 K

V2 = ?

Using the general gas equation = P1V1/T1 = P2V2/T2

V2 = P1V1T2 / P2T1

V2 = (741.45 × 43.60 × 273.15 ) / (760 × 293.15)

V2 = 39.63 ml

Volume at STP = 39.63 mL

d) 39.63 mL = 0.03963 L

Number of moles of oxygen gas in 0.03693 L = 0.03963/22.4 L

Number of molecules of oxygen gas = 0.03963/22.4 L × 6.03 × 10^23

Number of oxygen gas molecules = 1.065 × 10^21 molecules

e) Number of moles of O2 = 39.63/22.4 = 0.001769 mole

From the balanced equation, mole ratio between oxygen gas and potassium chlorate is 3 : 2

Moles KClO3 = 3/2 × 0.001769 moles = 0.00265 moles

Molar mass of KClO3 = 39 + 35.5 + 3 × 16 = 122.5 g

Actual mass of KClO3 decomposed = 122.5 grams × 0.00265 mole = 0.3246 grams

Percent purity = (actual mass KClO3 decomposed / sample mass of impure KClO3) × 100%

Percent purity = (0.3246/0.810) × 100% = 40.1 %

adefioyelaoye adefioyelaoye · Answer 2 · 2021-11-18T09:39:21+01:00

The balanced equation for the decomposition of potassium chlorate is

2KClO₃ ------> 2KCl + 3O₂

The corrected pressure of the dry oxygen gas will be calculated with this formula:

Vapor pressure of dry oxygen gas = Total pressure - (water vapor pressure + pressure due difference in water levels)

Vapor pressure of water at 20 °C is 17.535mm(torr)

Pressure due to different water levels is

4.22 cm × 10mm/cm / 13.534 (13.534 is the density of mercury) = 3.118 mm (torr).

Vapor pressure of dry oxygen gas = 762.10 torr - 3.118 torr - 17.535 torr

= 741.45 torr

Volume of dry oxygen gas

Since we have, P1 = 741.45 torr, V1 = 43.60 mL ,T1 = 20 °C + 273.15 = 293.15 K, P2 = 760 torr , T2 = 273.15 K , V2 = ?

We can then use the general gas equation = P1V1/T1 = P2V2/T2

V2 = P1V1T2 / P2T1

V2 = (741.45 × 43.60 × 273.15 ) / (760 × 293.15)

V2 = 39.63 mL

Volume of dry oxygen gas = 39.63 mL

Number of moles of oxygen gas that was collected will be 0.03693 L = 0.03963/22.4 L

Number of molecules of oxygen gas = 0.03963/22.4 L × 6.03 × 10^23

= 1.065 × 10^21 molecules

The percentage purity will be calculated using the number of moles of O₂ = 39.63/22.4 = 0.001769 mole

We use the balanced equation for the reaction to calculate mole ratio. The

mole ratio between oxygen gas and potassium chlorate is 3 : 2

Mole of KClO₃ = 3/2 × 0.001769 moles = 0.00265 moles

Molar mass of KClO₃ = 39 + 35.5 + 3 × 16 = 122.5 g

Actual mass of KClO₃ decomposed = 122.5 grams × 0.00265 mole = 0.3246 grams

Percentage purity = (actual mass KClO₃ decomposed / sample mass of impure KClO₃) × 100%

= (0.3246/0.810) × 100% = 40.1 %

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