what is the percent yield ... ?

[匿名]

2020-04-02 12:09 am

given the following reactions
Fe (s) + Br2 (g) --> FeBr2 (s)
3 FeBr2 (s) + Br2 (g) --> Fe3Br8 (s)

if each reaction gives an 82% yield, what mass of FeBr8 is produced from 1.00 g Fe, assuming Br2 is in excess?

please give the process too. thank you!

回答 (1)

andrew

2020-04-02 12:42 am

In the question, it should be "mass of Fe₃Br₈" instead of "mass of FeBr₈".

Molar mass of Fe = 55.85 g/mol
Molar mass of Fe₃Br₈ = (55.85×3 + 79.90×8) g/mol = 806.75 g/mol

3Fe(s) + 3Br₂(g) → 3FeBr₂(s)
3FeBr₂(s) + Br₂(g) → Fe₃Br₈(s)

Combine the above two equation, the overall equation becomes:
3Fe(s) + 4Br₂(g) → Fe₃Br₈(s)

Moles of Fe reacted = (1.00 g) / (55.85 g/mol) = 0.017905 mol
Maximum moles of Fe₃Br₈ produced = (0.017905 mol) × (1/3) = 0.005968 mol
Mass of Fe₃Br₈ produced = (0.005968 mol) × (806.75 g/mol) × 82% = 3.95 g

====
OR:
(1.00 g Fe) × (1 mol Fe / 55.85 g Fe) × (1 mol Fe₃Br₈ / 3 mol Fe) × (806.75 g Fe₃Br₈ / 1 mol Fe₃Br₈) × 82%
= 3.95 g Fe₃Br₈

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