Question #967d1

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Answer 1 · 2015-03-13T01:38:48

155.5g will be formed.

$4 F e + 3 O_{2} \to 2 F e_{2} O_{3}$ $4Fe+3O_2rarr2Fe_2O_3$

So 3 mol $O_{2}$ $O_2$ $\to$ $rarr$ 2 mol $F e_{2} O_{3}$ $Fe_2O_3$

Using approximate $A_{r}$ $A_r$ values:

$3 \times (2 \times 16)$ $3xx(2xx16)$ g $O_{2}$ $O_2$ $\to 2 [(56 \times 2) + (3 \times 16)]$ $rarr2[(56xx2)+(3xx16)]$ g $F e_{2} O_{3}$ $Fe_2O_3$

So:

$64 g \to 320 g$ $64grarr320g$

So:

$1 g \to \frac{320}{64} g$ $1grarr320/64g$

So:

$31.1 g \to \frac{320}{64} \times 31.1 = 155.5 g$ $31.1grarr(320)/(64)xx31.1=155.5g$