Question

Using the unbalanced reaction below:

UO2(s) + HF(aq) ----> UF4 + H2O(l)

How many grams of UF4 would we anticipate to get from 1.00 g of UO2 and an excess of HF

Answer 1

1.16 g of UF₄ can be formed in this reaction.

Explanation:

We have to write the balanced equation of the reaction between Uranium oxide and hydrogen fluoride to form Uranium tetra fluoride and water as,

UO₂(s) +4 HF(aq) → UF₄ + 4 H₂O(l)

We can find the moles of UF₄ and then multiplying molar mass of UF₄ with the moles of UF₄, we will get the mass of UF₄ in grams.

Moles of UF₄ =   $\frac{ 1 mol of UO_{2}\times 1 mol of UO_{2\times 1 mol of UF_{4} } }{270.03 g \times 1 mol of UO_{2} }$

                       = 0.00370 mol UF₄

Grams of UF₄ =0.00370 mol × 314.02 g/mol = 1.16 g of UF₄

So mass of UF₄ formed in the reaction is 1.16 g.