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

3 Answers
Jul 29, 2017

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........

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_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

Jul 29, 2017

"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".