What is the maximum mass of S8 that can be produced by combining 92.0 g of each reactant? 8SO2+16H2S⟶3S8+16H2O

Answer 1

Let's look at the balanced equation:

8SO₂ + 16H₂S ==> 3S₈ + 16H₂O

This tells us that 3 moles S₈ will be produced from 8 moles SO₂ and 16 moles H₂S. We must now determine how many moles of SO₂ and H₂S are actually present during this reaction and that will allow us to then calculate the maximum mass of S8 that can be produced. In other words, which reactant is present in limiting supply?

For SO₂: 92.0 g SO₂ x 1 mol SO₂/64.1g x 3 mol S₈/8 mol H₂S x 257 g S₈/mol S₈ = 138 g S₈ formed

For H₂S: 92.0 g H₂S x 1 mol H₂S/34.1 g x 3 mol S₈/16 mol H₂S x 257 g S₈/mol S₈ = 130. g S₈ formed

So, the maximum amount of S₈ that can be formed would be 130. g because the H₂S is limiting and thus limits the amount of product that can be produced.

Answer 2

Hello, Addison,

8SO2+16H2S⟶3S8+16H2O

Assuming the equation is balanced, we need to first find which (H2S or SO2) is the limiting reagent. We are given 92.0 grams (3sig figs) of both the SO2 and the H2S. We need to change those numbers into moles, since the mass alone doesn't allow us to compare their relative importance in the equation. We need the actual numbers of the molecules, and moles is easy the way to do that.

Molar masses (g/mole):

SO2 = 64.1

H2S = 34.1

To get moles, divided the given masses (92.0 grams each) by their respective molar masses:

moles SO2 = 92.0 grams/(64.1 g/mole) =1.44 moles SO2

moles H2S = 92.0 g H2S (34.1 g/mole) = 2.70 moles H2S

We see from the balanced equation that we need twice as many moles of H2S as SO2. I can tell from the numbers of actual moles that if we reacted ALL of the 1.44 moles of SO2, that we'd need twice that, or 2.88 moles of H2S. We only have 2.70 moles of H2S, however. Therefore, H2S will become the limiting reagent.

We'll assume that all of the H2S reacts with the excess of SO2, Since we start with 2.70 moles of H2S, the equation says we'll get 3/16 of that amount of S8 (3 moles of S8/18 moles of H2S).

So,

2.70 moles H2S*(3 moles S8/18 moles H2S) = 0.506 moles of S8.

The molar mass of S8 is 257 g/mole. Therefore, 0.506 moles S8 * (257 g/mole S8) = 130 grams S8

I hope this helps,

Bob