Question #824f3

Start by taking a look at the balanced chemical equation for this reaction

#color(red)(3)"Ag"_2"S"_text((s]) + color(blue)(2)"Al"_text((s]) + 3"H"_2"O"_text((l]) -> 6"Ag"_text((s]) + "Al"_2"O"_text(3(s]) + color(purple)(3)"H"_2"S"_text((aq])#

Your tool of choice for any stoichiometry problem is the mole ratio.

In this particular case, you have #color(red)(3)# moles of silver(I) sulfide reacting with #color(blue)(2)# moles of aluminium to form #color(purple)(3)# moles of hydrosulfuric acid.

This tells you that the two reactants will always mix in a #color(red)(3):color(blue)(2)# mole ratio, and the hydrosulfuric acid is always produced in a #color(red)(3):color(purple)(3)# mole ratio with silver(I) sulfide.

You know that you must remove #"125 g"# of silver(I) sulfide from the solution. Use the compound's molar mass to determine how many moles would be present in that sample

#125 color(red)(cancel(color(black)("g"))) * ("1 mole Ag"_2"S")/(247.8color(red)(cancel(color(black)("g")))) = "0.5044 moles Ag"_2"S"#

So, if this many moles of silver(I) sulfide are present in the sample, you can use the mole ratio that exists between the two reactants to determine how many moles of aluminium would be needed

#0.5044 color(red)(cancel(color(black)("moles Ag"_2"S"))) * (color(blue)(2)" moles Al")/(color(red)(3)color(red)(cancel(color(black)("moles Ag"_2"S")))) = "0.3363 moles Al"#

To get the mass of aluminium that would contain this many moles, use the element's molar mass

#0.3363 color(red)(cancel(color(black)("moles Al"))) * "26.98 g"/(1color(red)(cancel(color(black)("mole Al")))) = color(green)("9.07 g Al")#

The exact same approach can be used to find how many moles, and implicitly how many grams, of hydrosulfuric acid will be produced by this reaction.

Remember that you get equal numbers of moles of silver(I) sulfide and hydrosulfuric acid for this reaction, so you can say that

#n_(H_2S) = n_(Ag_2S) = "0.5044 moles"#

Now use hydrosulfuric acid's molar mass to find how many grams would contain this many moles

#0.5044 color(red)(cancel(color(black)("moles H"_2"S"))) * "34.08 g"/(1color(red)(cancel(color(black)("mole H"_2"S")))) = color(green)("17.2 g")#

The answers are rounded to three sig figs, the number of sig figs you have for the mass of silver(I) sulfide.