Question

2Fe(OH)3 Fe2O3+3H2O how many grams of Fe2O3 are produced if 10.7 grams of Fe (OH)3 react in this way

Answer 1

Answer: 7.98 grams of
`Fe_2O_3` are produced if 10.7 grams of
`Fe(OH)_3` are reacted.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Putting values in equation 1, we get:

`\text{Moles of} Fe(OH)_3=(10.7g)/(106.87g/mol)=0.100mol`

The chemical equation for the reaction is

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

By Stoichiometry of the reaction:

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

So, 0.100 moles of
`Fe(OH)_3` produce=
`(1)/(2)* 0.100=0.05mol` of
`Fe_2O_3`

Mass of
`Fe_2O_3` =
`moles* {\text{Molar Mass}}=0.05mol* 159.69g/mol=7.98g`

Hence 7.98 grams of
`Fe_2O_3` are produced if 10.7 grams of
`Fe(OH)_3` are reacted.