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Kazeer [188]
3 years ago
15

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

Chemistry
1 answer:
12345 [234]3 years ago
8 0

Answer:

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

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

<em>Given reaction</em>: 2S + 3O₂ → 2 SO₃

<em>In this reaction, the limiting reagent is S</em>

⇒ 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

⇒ <u><em>The amount of heat produced by 42.8 g S:</em></u>

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)

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

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Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

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