Question #15581

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Answer 1 · 2016-10-04T15:33:29

The formula of iron (III) oxide is $F e_{2} O_{3}$ $Fe_2O_3$ .

Considering atomic masses as

$F e \to 56 g/mol$ $Fe->56" g/mol"$

$O \to 16 g/mol$ $O->16" g/mol"$

Its formula mass

$= 2 \times 56 + 3 \times 16 = 160 g/mol$ $=2xx56+3xx16=160" g/mol"$

So 160 g Oxide contains 56 g Fe

Hence 100 g Iron(III) oxide will contain

$\frac{2 \times 56}{160} \times 100$ $(2xx56)/160xx100$ g Fe

$= 70$ $=70$ g Fe