Question

Fluorine react with uranium to produced aronian hexafluoride you UF6 as represented by this equation
U. + 3F2 =. UF6
How many fluorine molecules are required to produce 2.0 mg of uranium hexafluoride ,UF6, from an axis of uranium? The molar mass of UF6 is 352 gm/mol

Answer 1

Answer:

$1.0264\times 10^{19}$ fluorine molecules are required to produce 2.0 mg of uranium hexafluoride.

Explanation:

$U+ 3F_2\rightarrow UF_6$

Mass of uranium hexafluoride = 2.0 mg = 0.0020 g

Moles of uranium hexafluoride =$\frac{0.0020 g}{352 g/mol}=5.6818\times 10^{-6} moles$

According to reaction , 1 mol of uranium hexafluoride is obtained from 3 mole of fluorine gas.

Then $5.6818\times 10^{-6} moles$ of uranium hexafluoride will be obtained from:

$\frac{3}{1}\times 5.6818\times 10^{-6} mol=1.7045\times 10^{-5} moles$

Moles of fluorine gas required = $1.7045\times 10^{-5} moles$

Number of molecules of fluorine gas required:

$1.7045\times 10^{-5} mol\times N_A}$

$1.7045\times 10^{-5} mol\times 6.022\times 10^{23}mol^{-1}}=1.0264\times 10^{19} molecules$

$1.0264\times 10^{19}$ fluorine molecules are required to produce 2.0 mg of uranium hexafluoride.