Question

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

Answer 1

What does the stoichiometry say..........? It says that `1*mol` of hydrogen peroxide and `1*mol` hydrogen sulfide gives `2 *mol` of water and `1*mol` of sulfur.

Explanation:

You have the stoichiometric reaction:

`H_2O_2(l) + H_2S(g) rarr 2H_2O(l) + S(s)darr`

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*g` of hydrogen peroxide reacts with `34*g` hydrogen sulfide to give `36*g` water and `32*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*mol` hydrogen peroxide reacts, to give, THEREFORE, `27*g` `H_2O`, and `24*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*mol` `H_2O_2` represents a mass of `0.75*molxx34*g*mol^-1-=25.5*g`........etc........

Answer 2

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_2O_2+H_2Srarr2H_2O+S`

This equation tells us that for every mole of `H_2O_2`, two moles of water are produced. Therefore:

`n(H_2O)=2*n(H_2O_2)=2*0.75=1.5" "mol`

Answer 3

`"1.5 mol H"_2"O"`

Explanation:

Balanced Equation

`"H"_2"O"_2 + "H"_2"S"` `rarr` `"2H"_2"O" + "S"`

Multiply the given mol `"H"_2"O"` by the mole ratio between `"H"_2"O"` and `"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 `"H"_2"O"_2"` in the reactants, there are two moles `"H"_2"O"` in the products. So any number of moles of `"H"_2"O"_2"` will produce twice as many moles of `"H"_2"O"`.