Question

0.5 mole of FeS is made to react with 49 gram of H2SO4 the mass of h2s formed is

Answer 1

FeS + H2SO4 -------> H2S + FeSO4

1 mole H2SO4 = 98g
49g of H2SO4 = 0.5mole

From the reaction
1 mole FeS and 1 mole H2SO4 gives 1 mole H2S
So, 0.5 mole H2SO4 will give 0.5 mole H2S.

Answer 2

Answer: The mass of $H_2S$ formed will be 17 grams.

Explanation:

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

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

For sulfuric acid:

Given mass = 49 g

Molar mass = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{49g}{98g/mol}=0.5mol$

The chemical equation for the reaction of iron sulfide with sulfuric acid is given by:

$FeS+H_2SO_4\rightarrow FeSO_4+H_2S$

By Stoichiometry of the reaction:

1 mole of ferrous sulfide reacts with 1 mole of sulfuric acid.

So, 0.5 moles of ferrous sulfide will react with $\frac{1}{1}\times 0.5=0.5$ of sulfuric acid.

Hence, both the reactants will be completely consumed.

So, again by stoichiometry of the reaction:

1 mole of ferrous sulfide will produce 1 mole of hydrogen sulfide.

So, 0.5 moles of ferrous sulfide  will produce = $\frac{1}{1}\times 0.5=0.5mol$ of hydrogen sulfide.

• To calculate the amount of hydrogen sulfide, we use equation 1:

Moles of hydrogen sulfide = 0.5 moles

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

$0.5mol=\frac{\text{Mass of hydrogen sulfide}}{34g/mol}\\\\\text{Mass of hydrogen sulfide}=17g$

Hence, the mass of $H_2S$ formed will be 17 grams.