Question

How many grams of iron(III) oxide (Fe(OH)3) may theoretically be produced from 85.0 g FeCl3?

Answer 1

56.0 g

Step-by-step explanation:

Calculation of the moles of
`FeCl_3` as:-

Mass = 85.0 g

Molar mass of
`FeCl_3` = 162.2 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (85.0\ g)/(162.2\ g/mol)`

`Moles= 0.52404\ mol`

According to the reaction:-

`FeCl_3+3 NaOH\rightarrow Fe(OH)_3+3 NaCl`

1 mole of
`FeCl_3` on reaction forms 1 mole of
`Fe(OH)_3`

Also,

0.52404 mole of
`FeCl_3` on reaction forms 0.52404 mole of
`Fe(OH)_3`

Moles of
`Fe(OH)_3` = 0.52404 moles

Molar mass of
`Fe(OH)_3` = 106.867 g/mol

Mass = Moles*Molar mass =
`0.52404* 106.867\ g` = 56.0 g