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kondor19780726 [428]
2 years ago
13

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]2 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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Answer : Amoxicillin Suspension 125 mg/ 5 ml is 125 mg of Amoxicillin per 5 ml of suspension is an example of weight to volume.

Explanation :

Weight by volume (w/v) means that the mass of solute present in 100 mL volume of solution.

Weight by weight (w/w) means that the mass of solute present in 100 gram of solution.

Volume by volume (v/v) means that the volume of solute present in 100 mL volume of solution.

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Hence, it is an example of weight to volume.

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The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
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Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

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