Answer:
5.625 moles of oxygen, O₂.
Explanation:
The balanced equation for the reaction is given below:
4Al + 3O₂ —> 2Al₂O₃
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ required to react with 7.5 moles of aluminum, Al. This can be obtained as illustrated below:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Therefore, 7.5 moles of Al will react with = (7.5 × 3)/4 = 5.625 moles of O₂.
Thus, 5.625 moles of O₂ is needed for the reaction.
A) Head to tail joining of monomers. :) (confirmed correct answer, I took the test)
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The volume of H₂ evolved at NTP=0.336 L
<h3>Further explanation</h3>
Reaction
Decomposition of NH₃
2NH₃ ⇒ N₂ + 3H₂
conservation mass : mass reactants=mass product
0.28 NH₃= 0.25 N₂ + 0.03 H₂
2 g H₂ = 22.4 L
so for 0.03 g :
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