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

$\text{Moles of uranium}=\frac{1.00g}{238.03g/mol}=0.00420mol$

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 $0.00420\times 3=0.0126$ 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:

$0.980atm\times V=0.0126mole\times (0.0821L.atm/mol.K)\times 288K$

$V=0.304L$

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