Respuesta :
Answer:
4.86 moles of Al are needed to form 2.43 mol of Al2Br6.
Explanation:
[tex]2Al(s) + 3Br2(l)\rightarrow Al_2Br_6(s)[/tex]
Moles of [tex]Al_2Br_6=2.43 mol[/tex]
According to reaction, 1 mole of [tex]Al_2Br_6[/tex] is obtained from 2 moles of aluminum.
Then 2.43 mole of [tex]Al_2Br_6[/tex] will be obtained from :
[tex]\frac{2}{1}\times 2.43 mol=4.86 mol[/tex] of Al
4.86 moles of Al are needed to form 2.43 mol of [tex]Al_2Br_6[/tex].
4.86 moles of Al are needed to form 2.43 mol of Al2Br6.
Calculation for moles:
Balanced chemical reaction: 2 Al(s) + 3 Br₂(l) → Al₂Br₆ (s)
Given:
Moles of Al₂Br₆ = 2.43 moles
As per Stoichiometry, 1 mole of Al₂Br₆ is formed from 2 moles of aluminum.
Then 2.43 mole of will be obtained from : 2/1 * 2.43 = 4.86 moles of Al
4.86 moles of Al are needed to form 2.43 mol of Al₂Br₆ .
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