Question

What is the number of moles of ${\rm{Fe}}{\left( {{\rm{OH}}} \right)_{\rm{3}}}$
that can be produced by allowing 1 mole of ${\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}$, 2 moles of ${{\rm{H}}_{\rm{2}}}{\rm{O}}$ and 3 moles of ${{\rm{O}}_{\rm{2}}}$ to react? (as nearest integer)

${\rm{2F}}{{\rm{e}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}} + 6{{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{3}}{{\rm{O}}_2} \to 4{\rm{Fe}}{\left( {{\rm{OH}}} \right)_{\rm{3}}} + 6{\rm{S}}$

Answer 1

We know that a chemical reaction when written in balanced form gives a quantitative relationship in terms of molecules, masses, moles and volumes between different reactants and products involved in the reaction. This is called stoichiometry.

Complete step by step answer: Here, the chemical reaction is given as,
${\rm{2F}}{{\rm{e}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}} + 6{{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{3}}{{\rm{O}}_2} \to 4{\rm{Fe}}{\left( {{\rm{OH}}} \right)_{\rm{3}}} + 6{\rm{S}}$
…… (1)
Two moles of ${\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}$ reacts with six moles of water$\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)$ and 3 moles of oxygen $\left( {{\rm{oxygen}}} \right)$ to produce 4 moles of ferric hydroxide $\left( {{\rm{Fe}}{{\left( {{\rm{OH}}} \right)}_3}} \right)$ and three moles of sulphur.

We have to calculate the moles of ferric hydroxide produced when 1 mole of ${\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}$, 2 moles of ${{\rm{H}}_{\rm{2}}}{\rm{O}}$ and 3 moles of ${{\rm{O}}_{\rm{2}}}$ undergoes reaction.

So, from equation (1) we see that two moles of ${\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}$ requires 6 moles of ${{\rm{H}}_{\rm{2}}}{\rm{O}}$. So, one mole of ${\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}$ requires 3 moles of ${{\rm{H}}_{\rm{2}}}{\rm{O}}$. So, we can say that, water (${{\rm{H}}_{\rm{2}}}{\rm{O}}$) is the limiting reagent. So, the reaction will take place considering the amount of water (${{\rm{H}}_{\rm{2}}}{\rm{O}}$).

Now, we have to use the unitary method.
6 moles of water (${{\rm{H}}_{\rm{2}}}{\rm{O}}$) produces=4 moles of ferric hydroxide $\left( {{\rm{Fe}}{{\left( {{\rm{OH}}} \right)}_3}} \right)$

So, 2 moles of water produces ferric hydroxide=$\dfrac{4}{6} \times 2 = \dfrac{4}{3}\,{\rm{mol = 1}}{\rm{.33}}\,\,{\rm{mol}}$

Hence, the moles of ${\rm{Fe}}{\left( {{\rm{OH}}} \right)_3}$ formed when 1 mole of ${\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}$, 2 moles of ${{\rm{H}}_{\rm{2}}}{\rm{O}}$ and 3 moles of ${{\rm{O}}_{\rm{2}}}$ undergoes reaction is 1 mol (nearest integer).

Note: Remember that, the reactant which is present in lesser amounts in a chemical reaction is termed as limiting reactant or limiting reagent. It is called limiting reactant as it limits the participation of other reactants in the reaction. For example, in combustion of methane, methane is the limiting reactant because air is always available in excess. So, the amount of carbon dioxide and water formed in the reaction depends on the amount of methane not oxygen.