Question

1Kr + 3 F2 → 1KrF6

2. How many moles of fluorine are required to

produce 3.0 grams of KrFo?

Answer 1

Answer: 0.045 moles of
`F_2` will be required to produce 3.0 grams of
`KrF_6`.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} KrF_6=(3.0g)/(198g/mol)=0.015moles`

The balanced chemical equation is:

`Kr+3F_2\rightarrow KrF_6`

According to stoichiometry :

1 mole of
`KrF_6` is produced by = 3 moles of
`F_2`

Thus 0.015 moles of
`KrF_6` will be produced by=
`(3)/(1)* 0.015=0.045moles` of
`F_2`

Thus 0.045 moles of
`F_2` will be required to produce 3.0 grams of
`KrF_6`.