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3 years ago

How many grams of p2o5 will be obtained from the interaction of 11.2 liters of oxygen with phosphorus?

Chemistry

1 answer:

Amiraneli [1.4K]3 years ago

8 0

28.4 grams of P₂O₅ will be obtained from the interaction of 11.2 liters of oxygen with phosphorus.

Let's consider the following balanced equation.

4 P + 5 O₂ ⇒ 2 P₂O₅

Since the conditions are not specified, we will assume that we are working at standard temperature and pressure. At STP, 1 mole of an ideal gas occupies 22.4 L. The volume of 11.2 L of oxygen at STP, assuming ideal behavior, is:

$11.2 L \times \frac{1mol}{22.4L} = 0.500 mol$

The molar ratio of O₂ to P₂O₅ is 5:2. The moles of P₂O₅ obtained from 0.500 moles of O₂ are:

$0.500 mol O_2 \times \frac{2molP_2O_5}{5molO_2} =0.200 mol P_2O_5$

The molar mass of P₂O₅ is 141.94 g/mol. The mass corresponding to 0.200 moles of P₂O₅ is:

$0.200 mol \times \frac{141.94 g}{mol} = 28.4 g$

28.4 grams of P₂O₅ will be obtained from the interaction of 11.2 liters of oxygen with phosphorus.

You can learn more about stoichiometry here: brainly.com/question/22288091

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3 years ago

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3 years ago

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3 years ago

10. A small helium tank claims to be able to fill 30 balloons to a volume of 3.15 L

Sindrei [870]

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Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at constant temperature.

This law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; while if the pressure decreases, the volume increases.

Mathematically, Boyle's law states that the product of pressure and volume is constant:

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Studying two different states, an initial state 1 and a final state 2, it is satisfied:

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In this case, you know:

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Replacing in Boyle's law:

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Solving:

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In summary, 3.377 L of helium the tank will be able to produce at a pressure of 94.2 kPa.

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Explanation:

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3 years ago

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