Question

Determine the mass of Iron (III) oxide (Fe2O3) produced in the decomposition reaction of 100.0 grams of Iron (III) hydroxide (Fe(OH)3).

2Fe(OH)3(s)-1Fe2O3 (s) +3H2O (g)

Answer 1

To determine the mass of Iron (III) oxide produced in the decomposition reaction, use the stoichiometry of the balanced chemical equation and the given mass of Iron (III) hydroxide. First, calculate the moles of Fe(OH)3 using its molar mass. Then, use the stoichiometric mole ratio from the balanced equation to convert moles of Fe(OH)3 to moles of Fe2O3. Finally, multiply the moles of Fe2O3 by its molar mass to find the mass of Fe2O3 produced.

Step-by-step explanation:

To determine the mass of Iron (III) oxide produced in the decomposition reaction, we need to use the stoichiometry of the balanced chemical equation and the given mass of Iron (III) hydroxide. First, calculate the moles of Fe(OH)3 using its molar mass. Then, use the stoichiometric mole ratio from the balanced equation to convert moles of Fe(OH)3 to moles of Fe2O3. Finally, multiply the moles of Fe2O3 by its molar mass to find the mass of Fe2O3 produced.

Using the given balanced equation, we have:

2Fe(OH)3(s) → Fe2O3(s) + 3H2O(g)

Given mass of Fe(OH)3 = 100.0 g

Molar mass of Fe(OH)3 = 106.87 g/mol

Molar mass of Fe2O3 = 159.70 g/mol

First, calculate the moles of Fe(OH)3:

moles of Fe(OH)3 = mass of Fe(OH)3 / molar mass of Fe(OH)3

= 100.0 g / 106.87 g/mol

= 0.936 mol Fe(OH)3

Next, use the stoichiometric mole ratio from the balanced equation:

moles of Fe2O3 = moles of Fe(OH)3 × (1 mol Fe2O3 / 2 mol Fe(OH)3)

= 0.936 mol Fe(OH)3 × (1 mol Fe2O3 / 2 mol Fe(OH)3)

= 0.468 mol Fe2O3

Finally, calculate the mass of Fe2O3:

mass of Fe2O3 = moles of Fe2O3 × molar mass of Fe2O3

= 0.468 mol Fe2O3 × 159.70 g/mol

= 74.6 g Fe2O3

Answer 2

Answer: 77.47 g of
`Fe_2O_3` will be produced from 100.0 g of Iron (III) hydroxide

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Fe(OH)_3=(100.0g)/(106.9g/mol)=0.935moles`

The balanced chemical equation is:

`2Fe(OH)_3(s)\rightarrow Fe_2O_3(s)+3H_2O(g)`

According to stoichiometry :

2 moles of
`Fe(OH)_3` produce = 1 mole of
`Fe_2O_3`

Thus 0.935 moles of
`Fe(OH)_3` will producee=
`(1)/(2)* 0.935=0.468moles` of
`Fe_2O_3`

Mass of
`Fe_2O_3=moles* {\text {Molar mass}}=0.468moles* 159.7g/mol=74.74g`

Thus 77.47 g of
`Fe_2O_3` will be produced from 100.0 g of Iron (III) hydroxide