Question #2d232

Start with the balanced chemical equation for your reaction

#2"H"_2"S"_text((g]) + "SO"_text(2(g]) -> 3"S"_text((s]) + 2"H"_2"O"_text((l])#

Notice the mole ratios that exist between hydrogen sulfide and sulfur dioxide, on one hand, and sulfur on the other.

In order for the reaction to take place, for every mole of sulfur dioxide, you need 2 moles of hydrogen sulfide and produce 3 moles of solid sulfur.

Keep these ratios in mind.

Calculate the number of moles of each reactant by using their respective molar masses

#7.50color(red)(cancel(color(black)("g"))) * ("1 mole H"""_2"S")/(34.081color(red)(cancel(color(black)("g")))) = "0.2201 moles H"""_2"S"#

and

#12.75color(red)(cancel(color(black)("g"))) * ("1 mole SO"""_2)/(64.064color(red)(cancel(color(black)("g")))) = "0.1990 moles SO"""_2#

Since you need twice as many moles of hydrogen sulfide as you do of sulfur dioxide, it follows that hydrogen sulfide acts as a limiting reagent.

This means that the reaction will use up all the hydrogen sulfide, and leave you with excess sulfur dioxide.

If all the moles of hydrogen sulfide react, then the reaction will produce

#0.2201color(red)(cancel(color(black)("moles H"""_2"S"))) * "3 moles S"/(2color(red)(cancel(color(black)("moles H"""_2"S")))) = "0.3302 moles S"#

Use sulfur's molar mass to determine how many grams would contain this many moles

#0.3302color(red)(cancel(color(black)("moles"))) * "32.065 g"/(1color(red)(cancel(color(black)("mole S")))) = "10.588 g S"#

Rounded to three sig figs, the number of sig figs you have for the mass of hydrogen sulfide, the answer will be

#m_"S" = color(green)("10.6 g")#