How many moles of H2O could be obtained by reacting 0.75 mole of H2O2 in the reaction H2O2+H2S→2H2O+S?

Jul 29, 2017

What does the stoichiometry say..........? It says that $1 \cdot m o l$ of hydrogen peroxide and $1 \cdot m o l$ hydrogen sulfide gives $2 \cdot m o l$ of water and $1 \cdot m o l$ of sulfur.

Explanation:

You have the stoichiometric reaction:

${H}_{2} {O}_{2} \left(l\right) + {H}_{2} S \left(g\right) \rightarrow 2 {H}_{2} O \left(l\right) + S \left(s\right) \downarrow$

Is it balanced? For every reactant particle is there a corresponding product particle? There must be if the reaction represents chemical reality. It is balanced, and you have done the work not me. The reaction tells me UNEQUIVOCALLY that $34 \cdot g$ of hydrogen peroxide reacts with $34 \cdot g$ hydrogen sulfide to give $36 \cdot g$ water and $32 \cdot g$ sulfur. All of these masses correspond to molar equivalents. Charge and mass are balanced as required. From where did I get these masses? Did I just look them up?

Your starting conditions propose that $0.75 \cdot m o l$ hydrogen peroxide reacts, to give, THEREFORE, $27 \cdot g$ ${H}_{2} O$, and $24 \cdot g$ sulfur. Do you agree? This is an important principle to master, and if you don't from where we are coming, ask again.

And note that $0.75 \cdot m o l$ ${H}_{2} {O}_{2}$ represents a mass of $0.75 \cdot m o l \times 34 \cdot g \cdot m o {l}^{-} 1 \equiv 25.5 \cdot g$........etc........

Jul 29, 2017

1.5 moles of water could be obtained.

Explanation:

First, we always want the balanced chemical equation as this tells us the proportions that the reactants will react in, and how much of the products are formed.

${H}_{2} {O}_{2} + {H}_{2} S \rightarrow 2 {H}_{2} O + S$

This equation tells us that for every mole of ${H}_{2} {O}_{2}$, two moles of water are produced. Therefore:

$n \left({H}_{2} O\right) = 2 \cdot n \left({H}_{2} {O}_{2}\right) = 2 \cdot 0.75 = 1.5 \text{ } m o l$

Jul 29, 2017

$\text{1.5 mol H"_2"O}$

Explanation:

Balanced Equation

$\text{H"_2"O"_2 + "H"_2"S}$$\rightarrow$$\text{2H"_2"O" + "S}$

Multiply the given mol $\text{H"_2"O}$ by the mole ratio between $\text{H"_2"O}$ and ${\text{H"_2"O}}_{2}$.

0.75color(red)cancel(color(black)("mol H"_2"O"_2))xx(2"mol H"_2"O")/(1color(red)cancel(color(black)("mol H"_2"O"_2)))="1.5 mol H"_2"O"

You can reason out the answer without having to do the math. Notice by looking at the balanced equation, that for every one mole of $\text{H"_2"O"_2}$ in the reactants, there are two moles $\text{H"_2"O}$ in the products. So any number of moles of $\text{H"_2"O"_2}$ will produce twice as many moles of $\text{H"_2"O}$.