Question

What mass of phosphorus pentafluoride, pf5, has the same number of fluorine atoms as 25.0 g of oxygen difluoride, of2

Answer 1

The mass of phosphorus pentafluoride (PF5) that has the same number of fluorine atoms as 25.0 g of oxygen difluoride (OF2) is 43.8 g.

Step-by-step explanation:

The number of fluorine atoms in a molecule of phosphorus pentafluoride (PF5) is equal to the number of fluorine atoms in a molecule of oxygen difluoride (OF2). The molecular formula for PF5 indicates that there are 5 fluorine atoms. To find the mass of PF5 that has the same number of fluorine atoms as 25.0 g of OF2, we can use the concept of molar mass.

The molar mass of OF2 is 71.996 g/mol (16.00 g/mol for oxygen + 2(19.00 g/mol) for fluorine).

To find the mass of PF5, we need to know the molar mass. The molar mass of PF5 is 125.967 g/mol (30.97 g/mol for phosphorus + 5(19.00 g/mol) for fluorine).

Let's calculate the mass of PF5:

25.0 g OF2 × (125.967 g/mol / 71.996 g/mol) = 43.7997 g

Rounded to three significant figures: 43.8 g. Therefore, the mass of PF5 that has the same number of fluorine atoms as 25.0 g of OF2 is 43.8 g.

Answer 2

The mass of phosphorus pentafluoride
`(\(PF_5\))` that contains the same number of fluorine atoms as
`\(25.0 \, \text{g}\)` of oxygen difluoride
`(\(OF_2\))` remains approximately
`\(45.0 \, \text{g}\).`

To find the mass of phosphorus pentafluoride
`(\(PF_5\))` that has the same number of fluorine atoms as
`\(25.0 \, \text{g}\)` of oxygen difluoride
`(\(OF_2\))`.

The molar mass of oxygen difluoride
`(\(OF_2\)) is \(70.00 \, \text{g/mol}\)` (as previously calculated).

Given
`\(25.0 \, \text{g}\) of \(OF_2\)` and its molar mass, we can find the number of moles:

`\[\text{Number of moles of } OF_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{25.0 \, \text{g}}{70.00 \, \text{g/mol}}\]`

`\[\text{Number of moles of } OF_2 \approx 0.357 \, \text{mol}\]`

Now, phosphorus pentafluoride
`(\(PF_5\))` has 5 fluorine atoms per molecule.

The molar mass of phosphorus pentafluoride
`(\(PF_5\)) is \(125.97 \, \text{g/mol}\)` (as previously calculated).

To find the mass of
`\(PF_5\)` containing the same number of fluorine atoms as
`\(25.0 \, \text{g}\) of \(OF_2\)`, calculate the mass:

`\[\text{Mass of } PF_5 = \text{Number of moles of } OF_2 * \text{Molar mass of } PF_5\]`

`\[\text{Mass of } PF_5 = 0.357 \, \text{mol} * 125.97 \, \text{g/mol}\]`

`\[\text{Mass of } PF_5 \approx 45.0 \, \text{g}\]`