Question

Fluorine reacts with uranium to form $U{{F}_{6}}$
$U\left( s \right)+3{{F}_{2}}\left( g \right)\to U{{F}_{6}}\left( g \right)$
How many fluorine molecules are required to produce 2 mg of $U{{F}_{6}}$ from an excess of uranium? The molar mass of $U{{F}_{6}}$ is 352 g/mol.
A. $3.4\times {{10}^{38}}$
B. $1\times {{10}^{19}}$
C. $2\times {{10}^{19}}$
D. $3.4\times {{10}^{21}}$

Answer 1

We have to calculate the number of moles of the fluorine required to produce 2 mg of fluorine later we can convert the number of moles of fluorine into a number of molecules by multiplying with Avogadro number.
Avogadro number = $6.023\times {{10}^{23}}$ molecules or atoms.

Complete Solution :
- In the question it is given that Fluorine reacts with uranium to form $U{{F}_{6}}$.
- The chemical reaction to represent the above statement is as follows.
$U\left( s \right)+3{{F}_{2}}\left( g \right)\to U{{F}_{6}}\left( g \right)$

- In the above chemical reaction one mole of uranium reacts with three moles of fluorine to produce one mole of $U{{F}_{6}}$ as a product.
- Means 114 g of fluorine is required to produce 352 g of $U{{F}_{6}}$ .
- Therefore one gram of $U{{F}_{6}}$ is produced by

& = \dfrac{114}{352} \\\ & = 0.323g \\\ \end{aligned}$$ \- One gram of $U{{F}_{6}}$ is produced by 0.323 g of fluorine. \- To produce 2 mg of $U{{F}_{6}}$ we need = (0.323) (0.002) g = 0.000646 g of fluorine. \- The number moles of fluorine required is $$\begin{aligned} & =\dfrac{0.000646}{38} \\\ & =1.7\times {{10}^{-5}}mol \\\ \end{aligned}$$ \- The number of moles of fluorine present in 0.000646 g of fluorine is $1.7\times {{10}^{-5}}mol$ \- We know that one mole of fluorine contains Avogadro number of molecules ($6.023\times {{10}^{23}}$ ). \- Therefore $1.7\times {{10}^{-5}}mol$ contains number of fluorine molecules is $$\begin{aligned} & = 1.7\times {{10}^{-5}}\times 6.023\times {{10}^{23}} \\\ & = 1.0263\times {{10}^{19}}molecules \\\ \end{aligned}$$ \- Therefore $1.0263\times {{10}^{19}}$ molecules of fluorine is required to prepare 2mg of $U{{F}_{6}}$ . **So, the correct answer is “Option B”.** **Note:** First we calculated the number of moles of fluorine required to produce 2 mg of uranium hexafluoride and later we converted the number of moles of fluorine into molecules by multiplying with Avogadro number.