Question #15581

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Answer 1 · 2016-10-04T15:33:29

The formula of iron (III) oxide is $Fe_2O_3$ .

Considering atomic masses as

$Fe->56" g/mol"$

$O->16" g/mol"$

Its formula mass

$=2xx56+3xx16=160" g/mol"$

So 160 g Oxide contains 56 g Fe

Hence 100 g Iron(III) oxide will contain

$(2xx56)/160xx100$ g Fe

$=70$ g Fe