Question

What is the maximum amount (in grams) of sulfur dioxide that can be produced by the reaction of 1060 g of dihydrogen sulfide with 2890 g of oxygen according to the balanced equation below?2 H2S(g) + 3 O2(g) → 2 H2O(g) + 2 SO2(g)

Answer 1

The first thing to do when solving this problem is to make sure the equation is balanced:

$2H_2S_{(g)}+3O_{2(g)}\rightarrow2H_2O_{(g)}+2SO_{2(g)}$

We want to know the maximum amount of SO2 that can be produced.

Given:

mass of H2S = 1060 g

mass of O2 = 2890 g

We have to find the limiting reagent, and use its molar ratio to find number of moles of SO2, then we can find the mass produced.

H2S:

n = m/M where n is the number of moles, m is the mass and M is the molar mass. Molar mass can be found by adding atomic masses of H2 + S.

n = 1060 g/34,1 g/mol

n = 31.085 mol

O2:

n = m/M

n = 2890 g/31.998

n = 90.318 mol

So Oxygen has a higher number of moles, so it is reactant in excess and H2S the limiting reactant/reagent since it has less number of moles.

We can use the stoichiometry to find the molar ratios of H2S:SO2. The molar ratio is 2:2.

Therefore the number of moles of SO2 = 31.085 x (2/2) = 31.085 mol

From the number of moles, we can find the mass of SO2.

m = n x M

m = 31.085 mol x 64,066 g/mol

m = 1991.5 g of sulfur dioxide will be produced.