Answer: 0.045 moles of [tex]F_2[/tex] will be required to produce 3.0 grams of [tex]KrF_6[/tex].
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} KrF_6=\frac{3.0g}{198g/mol}=0.015moles[/tex]
The balanced chemical equation is:
[tex]Kr+3F_2\rightarrow KrF_6[/tex]
According to stoichiometry :
1 mole of [tex]KrF_6[/tex] is produced by = 3 moles of [tex]F_2[/tex]
Thus 0.015 moles of [tex]KrF_6[/tex] will be produced by=[tex]\frac{3}{1}\times 0.015=0.045moles[/tex] of [tex]F_2[/tex]
Thus 0.045 moles of [tex]F_2[/tex] will be required to produce 3.0 grams of [tex]KrF_6[/tex].
