Question

What amount of iron sulphide is formed when 11.2 g of iron is heated with sulphur?
Fe + S + Fes​

Answer 1

The amount of iron sulfide present is 17.6 grams

Explanation:

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

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

Given mass of iron = 11.2 g

Molar mass of iron = 55.85 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron}=\frac{11.2g}{55.85g/mol}=0.200mol$

For the given chemical equation:

$Fe+S\rightarrow FeS$

By Stoichiometry of the reaction:

1 mole of iron produces 1 mole of iron sulfide

So, 0.200 moles of iron will produce = $\frac{1}{1}\times 0.200=0.200mol$ of iron sulfide

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

Molar mass of iron sulfide = 88 g/mol

Moles of iron sulfide = 0.200 moles

Putting values in equation 1, we get:

$0.200mol=\frac{\text{Mass of iron sulfide}}{88g/mol}\\\\\text{Mass of iron sulfide}=(0.200mol\times 88g/mol)=17.6g$

Learn more about number of moles and stoichiometry:

https://brainly.com/question/919137

https://brainly.in/question/13443844

#learnwithbrainly