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)}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7Bmass%5C%3A%20%28w_%7B1%7D%29%7D%7Bmolar%5C%3A%20mass%5C%3A%20%28m%29%7D)
∴ mass of S in 2 moles of S: ![w_{1} = n \times m = 2\: mol \times 32\: g/mol = 64\: g](https://tex.z-dn.net/?f=w_%7B1%7D%20%3D%20n%20%5Ctimes%20m%20%3D%202%5C%3A%20mol%20%5Ctimes%2032%5C%3A%20g%2Fmol%20%3D%2064%5C%3A%20g)
<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](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B%28-%20792%5C%3A%20kJ%29%20%5Ctimes%2042.8%5C%3A%20g%7D%7B64%5C%3A%20g%7D%20%3D%20%28-529.65%29%5C%3A%20kJ)
![\Rightarrow x = (-5.2965 \times 10^{2})\: kJ = (-5.2965 \times 10^{5})\: J](https://tex.z-dn.net/?f=%5CRightarrow%20x%20%3D%20%28-5.2965%20%5Ctimes%2010%5E%7B2%7D%29%5C%3A%20kJ%20%3D%20%28-5.2965%20%5Ctimes%2010%5E%7B5%7D%29%5C%3A%20J)
![(\because 1 kJ = 10^{3} J)](https://tex.z-dn.net/?f=%28%5Cbecause%201%20kJ%20%3D%2010%5E%7B3%7D%20J%29)
<u>Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>