Question

What volume of F2 (in liters) is required to react with 1.00 g of uranium according to the equation U(s) + 3F2(g) → UF6(g), if the temperature is 15°C and the pressure is 745 mm Hg?

Answer 1

Answer : The volume of
`F_2` required is, 0.304 L

Explanation :

First we have to calculate the moles of uranium.

`\text{Moles of uranium}=\frac{\text{Mass of uranium}}{\text{Molar mass of uranium}}`

Molar mass of uranium = 238.03 g/mol

`\text{Moles of uranium}=(1.00g)/(238.03g/mol)=0.00420mol`

Now we have to calculate the moles of
`F_2`

The given balanced chemical reaction is:

`U(s)+3F_2(g)\rightarrow UF_6(g)`

From the balanced chemical reaction we conclude that,

As, 1 mole of uranium react with 3 moles of
`F_2`

So, 0.00420 mole of uranium react with
`0.00420* 3=0.0126` moles of
`F_2`

Now we have to calculate the volume of
`F_2`

Using ideal gas equation:

`PV=nRT`

where,

P = Pressure of
`F_2` gas = 745 mmHg = 0.980 atm (1 atm = 760 mmHg)

V = Volume of
`F_2` gas = ?

n = number of moles
`F_2` = 0.0126 mole

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of
`F_2` gas =
`15^oC=273+15=288K`

Putting values in above equation, we get:

`0.980atm* V=0.0126mole* (0.0821L.atm/mol.K)* 288K`

`V=0.304L`

Thus, the volume of
`F_2` required is, 0.304 L