Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation:

2 s + 3 o2 → 2 so3 δh = - 792 kj

if 42.8 g of s is reacted with excess o2, what will be the amount of heat given off?

Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J

Explanation:

Given reaction: 2S + 3O₂ → 2 SO₃

Given: The enthalpy of reaction: ΔH = - 792 kJ

Given mass of S: w₂ = 42.8 g, Molar mass of S: m = 32 g/mol

In the given reaction, the number of moles of S reacting: n = 2

As, Number of moles:

∴  mass of S in 2 moles of S:

Given reaction: 2S + 3O₂ → 2 SO₃

In this reaction, the limiting reagent is S

⇒ 2 moles S produces (- 792 kJ) heat.

or, 64 g of S produces (- 792 kJ) heat.

∴ 42.8 g of S produces (x) amount of heat

⇒ The amount of heat produced by 42.8 g S:

Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J

56 moles of   lieter in chemistry