P₄0₆
+ 20₂
⇒ P₄0₁₀
Explanation:
The overall equation for the reaction that produces P₄0₁₀ is :
P₄0₆ + 20₂ ⇒ P₄0₁₀
Now let us derive this equation:
Given equations:
P₄ + 30₂ ⇒ P₄0₆ equation 1;
P₄ + 50₂ ⇒ P₄0₁₀ equation 2;
To get the overall combined equation, the equation 1 must be reversed and added to equation 2:
P₄0₆ ⇒ P₄ + 30₂ equation 3
+
equation 2:
P₄ + 50₂ + P₄0₆ ⇒ P₄0₁₀ + P₄ + 30₂
cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;
P₄0₆ + 20₂ ⇒ P₄0₁₀
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Answer:
Explanation:
2S + 3O₂ = 2SO₃
2moles 3 moles
2 moles of S react with 3 moles of O₂
5 moles of S will react with 3 x 5 / 2 moles of O₂
= 7.5 moles of O₂ .
O₂ remaining unreacted = 10 - 7.5 = 2.5 moles .
All the moles of S will exhausted in the reaction and 2.5 moles of oxygen will be left .
its me again this is how you find the answer. note: i dont have the periodic table to see the exact atomic mass to find the molar mass however this is the answer:
so the actual formula is Mg(OH) with a 2 as a subscript because there are 2 Mg. so with Mg(OH)2 the Molar mass is 58.32g/mol. 58.32 g/mol x 7.1x1024 = 4.1x1026g
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
