Question

Flourine reacts with uranium to produce uranium hexafluoride ($U{F}_6$) as represented by this equation:
$U\left(S\right)+3F_2\left(g\right)\to U{F}_6\left(g\right)$
Number of fluorine molecules required to produce $7.04$ mg of uranium hexafluoride ($U{F}_6$) from an excess of uranium is :
(Molar mass of $U{F}_6$ is $352$ g/mol and $N_A=6\times {10}^{23}$)

• A. $3.6\times {10}^{19}$
• B. $3.6\times {10}^{15}$
• C. $4.6\times {10}^{19}$
• D. $4.6\times {10}^{15}$

Answer 1

Answer: (A)

$3.6\times {10}^{19}$

$7.04$ mg of uranium hexafluoride corresponds to $\frac{7.04}{1000\times 352}=2\times {10}^{-5}$ moles or $2\times {10}^{-5}\times 6.023\times {10}^{23}=1.2\times {10}^{19}$ molecules.
The number of fluorine molecules required is $3\times 1.2\times {10}^{19}=3.6\times {10}^{19}$.