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

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........

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#

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"#.