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3 years ago

What is the volume (in dm3 of 1 mole of oxygen at 5 MPa and 200 K?

Chemistry

1 answer:

vodka [1.7K]3 years ago

7 0

Answer: The volume of oxygen gas is $0.332dm^3$

Explanation:

To calculate the volume of the gas, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = pressure of the gas = 5 MPa = 5000 kPa (Conversion factor: 1 MPa = 1000 kPa)

V = Volume of gas = 3.34 L

n = number of moles of oxygen gas = 1 mole

R = Gas constant = $8.31dm^3\text{ kPa }mol^{-1}K^{-1}$

T = Temperature of the gas = 200 K

Putting values in above equation, we get:

$5000kPa\times V=1mol\times 8.31dm^3\text{ kPa }mol^{-1}K^{-1}\times 200K\\\\V=0.332dm^3$

Hence, the volume of oxygen gas is $0.332dm^3$

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How many moles of water are produced if 5.43 mol PbO2 are consumed? ANSWER: 10.9

Arada [10]

Answer: 10.9 mol.

Explanation:

To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.

Using cross multiplication:

1.0 mole of PbO₂ → 2.0 moles of H₂O

5.43 moles of PbO₂ → ??? moles of water

The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.

4 0

3 years ago

Read 2 more answers

Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.

weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100