Question

One of the first steps in the enrichment of uranium for use in nuclear power plants involves a displacement reaction between ${\mathbf{UO}}_{\mathbf{2}}\mathbf{?}$ and aqueous HF:

${\mathbf{UO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{HF}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }{\mathbf{UF}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\left(}\mathbf{unbalanced}\mathbf{\right)}$

How many liters of 2.40 M HF will react with 2.15 kg of${\mathbf{UO}}_{\mathbf{2}}\mathbf{?}$

Short answer

13.27 lit liters of 2.40 M HF will react with 2.15 kg of${\mathrm{UO}}_2$.

Step-by-step solution

Write down the balanced molecular equation

The reaction is shown below.

${\mathrm{UO}}_2\left(\mathrm{s}\right)+4\mathrm{HF}\left(\mathrm{aq}\right)\to {\mathrm{UF}}_4\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Calculate the moles of HF

Given values are:

The mass of = 2.15 kg and the molarity of HF = 2.40 M

Moles of HFare:

$\begin{array}{c}{\text{number\hspace{0.17em}of\hspace{0.17em}moles\hspace{0.17em}of\hspace{0.17em}UO}}_{\text{2}}=\frac{\text{2.15\hspace{0.17em}kg×1000\hspace{0.17em}g}}{\text{270.03}}\\ =7.96\text{\hspace{0.17em}moles}\end{array}$

One mole of ${\mathrm{UO}}_2$react with 4 moles of HF. Therefore,

$\text{mole\hspace{0.17em}\hspace{0.17em}ratio\hspace{0.17em}=\hspace{0.17em}}\frac{\text{4\hspace{0.17em}\hspace{0.17em}moles\hspace{0.17em}\hspace{0.17em}HF}}{{\text{1\hspace{0.17em}\hspace{0.17em}moleUO}}_{\text{2}}}=4$

Number of moles of HF that reacted with$7.96\text{\hspace{0.17em}moles}$ of${\mathrm{UO}}_2$ is calculated below.

$\text{Number\hspace{0.17em}of\hspace{0.17em}molesof\hspace{0.17em}HF}=7.96\times 4=31.85$

Calculate the volume of HF

The volume of HF to react completely with 2.15 kg of${\mathrm{UO}}_2$

role="math" localid="1658752947679" $\begin{array}{ccc}\mathrm{volume}\mathrm{of}\mathrm{HF}& =& \frac{\mathrm{moles}\mathrm{of}\mathrm{HF}}{\mathrm{molarity}\mathrm{of}\mathrm{HF}}\\ & =& \frac{31.85\mathrm{molHF}}{\left(\frac{2.40\mathrm{molHF}}{1\mathrm{litHF}}\right)}\\ & & =13.27\mathrm{literHF}\end{array}$