Question

In the following reaction, how many grams of ferrous sulfide (FeS) will produce

0.56 grams of iron (III) oxide (Fe2O3)?
4FeS + 7O2 → 2Fe2O3 + 4SO2
The molar mass of ferrous sulfide is 87.92 grams and that of iron (III) oxide is 159.7 grams.

Answer 1

First find no. of moles of Fe2O3 formed
next find the no. of moles of FeS by mole ratio
next find the mass of FeS
mass of FeS = 0.617g

Answer 2

Answer: The mass of ferrous sulfide required is 0.615 g

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)

• For iron (III) oxide:

Given mass of iron (III) oxide = 0.56 g

Molar mass of iron (III) oxide = 159.7 g/mol

Putting values in equation 1, we get:

`\text{Moles of iron (III) oxide}=(0.56g)/(159.7g/mol)=0.0035mol`

For the given chemical reaction:

`4FeS+7O_2\rightarrow 2Fe_2O_3+4SO_2`

By Stoichiometry of the reaction:

2 moles of iron (III) oxide is produced from 4 moles of ferrous sulfide.

So, 0.0035 moles of iron (III) oxide will be produced from =
`(4)/(2)* 0.0035=0.007mol` of ferrous sulfide.

Now, calculating the mass of ferrous sulfide from equation 1, we get:

Molar mass of ferrous sulfide = 87.92 g/mol

Moles of ferrous sulfide = 0.007 moles

Putting values in equation 1, we get:

`0.007mol=\frac{\text{Mass of ferrous sulfide}}{87.92g/mol}\\\\\text{Mass of ferrous sulfide}=0.615g`

Hence, the mass of ferrous sulfide required is 0.615 g