Question

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

Answer 1

Answer:

56.0 g

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{85.0\ g}{162.2\ g/mol}$

$Moles= 0.52404\ mol$

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 = $0.52404\times 106.867\ g$ = 56.0 g