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grigory [225]
3 years ago
8

Given that 2S (s)+3O2 (g)→2SO3 (g)2SO2 (g)+O2 (g)→2SO3 (g) has an enthalpy change of −790.4 kJ has an enthalpy change of −198.2

kJ. What is the heat of formation of SO2 in kilojoules?S (s)+O2 (g)→SO2 (g)?
Chemistry
1 answer:
abruzzese [7]3 years ago
4 0

Answer:

The heat of formation of SO2 is -296.1 kJ

Explanation:

<u>Step 1:</u> Data given

2S (s)+3O2 (g)→2SO3 (g)     ΔH = -790.4 kJ  

2SO2 (g)+O2 (g)→2SO3 (g)  ΔH = -198.2 kJ

<u>Step 2</u>: Calculate the heat of formation of SO2

2 S(s) + 3 O2(g) --> 2 SO3(g) ΔH = -790.4 kJ  

S(s) + 3/2 O2(g) → SO3(g)    ΔH = -395.2 kJ

2SO2 (g)+O2 (g)→2SO3 (g)  ΔH = -198.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g)   ΔH = 99.1 kJ

----------------------------------------------------------------

S(s) + 3/2 O2(g) → SO3(g)    ΔH = -395.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g)   ΔH = 99.1 kJ

-------------------------------------------------------------------

S (s)+O2 (g)→SO2 (g)

ΔHrxn = (-790.4 /2) kJ + (198.2/2) kJ

ΔHrxn = -395.2 kJ + 99.1 kJ = 296.1 kJ

The heat of formation of SO2 is -296.1 kJ

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kari74 [83]

Answer:

177.3kg C₂₁H₄₄

Explanation:

Based on the chemical reaction:

C₂₁H₄₄ → 3C₂H₄ + C₁₅H₃₂

<em>Where 1 mole of C₂₁H₄₄ produce 3 moles of ethene, C₂H₄.</em>

<em />

To solve this question we need to determine the moles of ethene in 50.4kg. 1/3 these moles are the moles of C₂₁H₄₄ that must be added:

<em>Moles Ethene -Molar mass: 28.05g/mol-</em>

50.4kg = 50400g * (1mol / 28.05g) = 1796.8 moles of ethene

<em>Moles C₂₁H₄₄:</em>

1796.8 moles of ethene * (1 mol C₂₁H₄₄ / 3 mol C₂H₄) = 589.93 moles C₂₁H₄₄

<em>Mass C₂₁H₄₄:</em>

589.93 moles C₂₁H₄₄ * (296g / mol) = 177283g =

<h3>177.3kg C₂₁H₄₄</h3>
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