Question

How many grams of $NaBr$ is produced if 55.85 g of $Fe$ is consumed in the following series of reactions assuming that $B{r}_2$ and $N{a}_{2}C{O}_3$ are in excess?
(molar masses of $Na,Br$ and $Fe$ are respectively $23,80$ and $55.85{\text{g mol}}^{-1})$
$3Fe+3B{r}_2\to 3FeB{r}_2$
$3FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8$
$F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3\to 8NaBr+4C{O}_2+F{e}_3{O}_4$

Answer 1

$3Fe+3B{r}_2\to 3FeB{r}_2\left(i\right)$ $3FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8\left(ii\right)$
$F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3\to 8NaBr+4C{O}_2+F{e}_3{O}_4\left(iii\right)$

Mol of Fe = 1 mol (given)
In reaction (i), 3 mol of Fe produce 3 mol of $FeB{r}_2$.
hence, 1 mol will produce 1 mol of $FeB{r}_2$.

In (ii) reaction, 3 mol of $FeB{r}_2$ produces 1 mol of $F{e}_{3}B{r}_8$.
So, 1 mol will produce 0.33 mol.

In (iii) reaction, 1 mol of $F{e}_{3}B{r}_8$ produces 8 mol of $NaBr$.
So, 0.33 mol produce = 8 $\times$ 0.33 = 2.64 mol of $NaBr$.
Amount of $NaBr$ in grams = 2.64 mol $\times$ 103 g/mol = 271.92 g