Question

How many moles of UF6 would have to be decomposed to provide enough fluorine to prepare 8.99 mol of CF4? (Assume sufficient carbon is available).

Answer 1

Answer:

5.99 moles of $UF_6$

Explanation:

In this case, we can start with the decomposition of $UF_6$, so:

$UF_6~->~U~+~3F_2$ (Reaction 1)

The $F_2$ can react with carbon to produce $CF_4$:

$CF_4~+~2F_2~->~CF_4$ (Reaction 2)

If we 8.99 mol of $CF_4$, we can calculate the moles of $F_2$ that we need. In reaction 2 we have a molar ratio of 1:2 (2 moles of $F_2$ will produce 1 mol of $CF_4$):

$8.99~mol~CF_4\frac{2~mol~F_2}{1~mol~CF_4}~=~17.98~mol~F_2$

With this value and using the molar ratio in reaction 1 (3 moles of $F_2$ are producing by each mol of $UF_6$), so:

$17.98~mol~F_2\frac{1~mol~UF_6}{3~mol~F_2}~=~5.99~mol~UF_6$

So, we will need 5.99 moles of $UF_6$ to produce 8.99 mol of $CF_4$.

I hope it helps!