Question

A 125 g sample of iron (III) bromide, FeBr3, is weighed out on a balance. (a) How many moles of the compound are present? (b) How many iron (III) ions are present in the sample? (c) How many ions of bromide are present in the sample? An excellent response will clearly show each step in your reasoning, indicating the units (dimensions) of the answers and consideration of significant figures. For answers greater than 1000 or less than 0.1 scientific notation should be used.

Answer 1

a)0.4222 moles of the compound are present.

b)
`2.542* 10^(23) ions` of iron(III)

c)
`7.6274* 10^(23) ions` of bromide.

Step-by-step explanation:

Moles =
`\frac{\text[Mass of compound}}{\text{Molar mas of compound}}`

a)Moles of iron (III) bromide =
`(125 g)/(296 g/mol)=0.4222 mol`

0.4222 moles of the compound are present.

Molecules of of iron (III) bromide in 0.4222 moles

1 mole =
`6.022* 10^(23)` molecules

In , 0.4222 moles:

`6.022* 10^(23)* 0.4222 molecules`

`=2.542* 10^(23) molecules` of iron (III) bromide

b)1 molecule of iron (III) bromide contain 1 iron(III) ion.Then
`2.542* 10^(23) molecules` will contain:

`1* 2.542* 10^(23) molecule= 2.542* 10^(23) ions` of iron(III)

c) 1 molecule of iron (III) bromide contain 3 bromide ion.Then
`2.542* 10^(23) molecules` will contain:

`3* 2.542* 10^(23) molecule= 7.6274* 10^(23) ions` of bromide.