Question

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?

Answer 1

Answer:

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: $n = \frac{mass\: (w_{1})}{molar\: mass\: (m)}$

∴  mass of S in 2 moles of S: $w_{1} = n \times m = 2\: mol \times 32\: g/mol = 64\: g$

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:

$x = \frac{(- 792\: kJ) \times 42.8\: g}{64\: g} = (-529.65)\: kJ$

$\Rightarrow x = (-5.2965 \times 10^{2})\: kJ = (-5.2965 \times 10^{5})\: J$

$(\because 1 kJ = 10^{3} J)$

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