Question #1ff7e

1 Answer
May 24, 2017

Answer:

A 4.50-mole sample of #"SnF"_2"# will contain 171 g of fluoride ions.

Explanation:

The formula unit for the ionic compound stannous fluoride #("SnF"_2)# indicates that there are two moles of fluoride ions #("F"^(-))# in one mole of stannous fluoride.

#4.50color(red)cancel(color(black)("mol SnF"_2))xx(2"mol F"^(-))/(1color(red)cancel(color(black)("mol SnF"_2)))="9.00 mol F"^(-)#

Multiply the mol #"F"^(-)# by its molar mass: #"18.998 g/mol"#. (The atomic weight of F to five sig figs on the periodic table in g/mol.)

#9.00color(red)cancel(color(black)("mol F"^(-)))xx(18.998"g F"^(-))/(1color(red)cancel(color(black)("mol F"^(-))))="171 g F"^(-)# (rounded to three sig figs)