Question

One of the first steps in the enrichment of uranium for use in nuclear power plants involves a displacement reaction between UO2 and aqueous HF: UO2(s) + HF(aq) → UF4(s) + H2O(l) [unbalanced] How many liters of 2.27 M HF will react with 3.70 kg of UO2?

Answer 1

Answer: 24.1 L of 2.27 M HF will react with 3.70 kg of $UO_2$

Explanation:

The balanced chemical reaction :

$UO_2(s)+4HF(aq)\rightarrow UF_4(s)+2H_2O(l)$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of } UO_2=\frac{3.70\times 1000g}{270g/mol}=13.7moles$

According to stoichiometry :

1 mole of $UO_2$ require = 4 moles of $HF$

Thus 13.7 moles of $UO_2$ will require=$\frac{4}{1}\times 13.7=54.8moles$  of $HF$

To calculate the volume for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution in L)}}$     .....(1)

Molarity of $HF$ solution = 2.27 M

Volume of $HF$ solution in L= ?

Putting values in equation 1, we get:

$54.8moles={2.27M}\times V}$

$V=24.1L$

Thus 24.1 L of 2.27 M HF will react with 3.70 kg of $UO_2$