What is the maximum mass of S8 that can be produced by combining 88.0 g of each reactant? 8SO2+16H2S⟶3S8+16H2O

124.24 g of S₈.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

8SO₂ + 16H₂S —> 3S₈ + 16H₂O

Next, we shall determine the masses of SO₂ and H₂S that reacted and the mass of S₈ produced from the balanced equation.

This is illustrated below:

Molar mass of SO₂ = 32 + (16×2)

= 32 + 32 = 64 g/mol

Mass of SO₂ from the balanced equation = 8 × 64 = 512 g

Molar mass of H₂S = (2×1) + 32

= 2 + 32 = 34 g/mol

Mass of H₂S from the balanced equation = 16 × 34 = 544 g

Molar mass of S₈ = 32 × 8 = 256 g/mol

Mass of S₈ from the balanced equation = 3 × 256 = 768 g

Thus,

From the balanced equation above,

512 g of SO₂ reacted with 544 g of H₂S to produce 768 g of S₈.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

512 g of SO₂ reacted with 544 g of H₂S.

Therefore, 88 g of SO₂ will react with = (88 × 544) / 512 = 93.5 g of H₂S.

From the calculation made above, we can see that it will take a higher mass (i.e 93.5 g) than what was given (i.e 88g) of H₂S to react completely with 88 g of SO₂.

Therefore, H₂S is the limiting reactant and SO₂ is the excess reactant.

Finally, we shall determine the maximum mass of S₈ produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of S₈ since all of it is consumed in the reaction.

The limiting reactant is H₂S and the maximum mass of S₈ produced can be obtained as follow:

From the balanced equation above,

544 g of H₂S reacted to produce 768 g of S₈.

Therefore, 88 g of H₂S will react to produce = (88 × 768) / 544 = 124.24 g of S₈.

Therefore, the maximum mass of S₈ produced from the reaction is 124.24 g.