Hydrogen peroxide was catalytically decomposed and 75.3 mL of oxygen gas was collected over water at 25°C and 742 torr. What mass of oxygen was collec

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02-03-2020
Chemistry

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Hydrogen peroxide was catalytically decomposed and 75.3 mL of oxygen gas was collected over water at 25°C and 742 torr. What mass of oxygen was collec

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Tringa0 Tringa0 · Answer 1 · 2020-03-03T10:24:23+01:00

Answer:

0.0928 grams of oxygen was collected.

Explanation:

Volume of oxygen gas collected = V= 75.3 mL = 0.0753 L (1 mL = 0.001 L)

Temperature of the gas = T = 25°C = 25+ 273 k = 298 K

Pressure of the gas = P - [tex]P_w[/tex]= 742 Torr - 24 Torr = 718 Torr

1 atm = 760 Torr

[tex]718 Torr=\frac{718}{760} atm=0.945 atm[/tex]

Moles of oxygen gas = n

Using an ideal gas equation;

[tex]PV=nRT\\n=\frac{PV}{RT}[/tex]

[tex]n=\frac{0.945 atm\times 0.0753 L}{0.0821 atm L/mol K\times 298 K}[/tex]

n = 0.0029 mol

Mass of 0.0029 moles of oxygen gas :

0.0029 mol × 32 g/mol = 0.0928 g

0.0928 grams of oxygen was collected.

ajeigbeibraheem ajeigbeibraheem · Answer 2 · 2021-11-30T13:32:35+01:00

The mass of the oxygen collected is 0.0930 grams.

From the given information;

the volume of the oxygen = 75.3 mL
the temperature = 25° C
the total partial pressure = 742 torr

Assuming that at a temperature of 25° C, the partial pressure of water = 24 torr

Then, according to Dalton law of Partial pressure law:

[tex]\mathbf{P_{total }= P_{O_2}+P_{water}}[/tex]

[tex]\mathbf{742 torr = P_{O_2}+24 \ torr}[/tex]

[tex]\mathbf{ P_{O_2} = 742\ torr-24 \ torr}[/tex]

[tex]\mathbf{ P_{O_2} = 718 \ torr}[/tex]

Now, we need to convert the pressure of Oxygen from torr to atmospheric pressure (atm)

At standard conversion rate;

1 torr = 0.00131579 atm

∴

718 torr will be = (718 torr × 0.00131579 atm)/ 1 torr

= 0.9447 atm

By using the ideal gas equation, we can determine the mass of the oxygen collected.

We know that:

PV = nRT

Making (n), the subject of the formula, we have:

[tex]\mathbf{n = \dfrac{PV}{RT}}[/tex]

[tex]\mathbf{n = \dfrac{0.9447 atm \times 75.3 \times 10 ^{-3} \ L }{(0.0821 \ L.atm .mol^{-1}K^{-1} \times 298 K }}[/tex]

n = 0.002907 moles

From the relation of number of moles, recall that:

number of moles = mass / molar mass

∴

mass = number of moles × molar mass

molar mass of oxygen = 32 g/mol

∴

mass = 0.002907 moles × 32 g/mol
mass of oxygen = 0.0930 grams

Therefore, we can conclude that the mass of oxygen collected was 0.0930 grams.

Learn more about Dalton's law of partial pressure here:

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