# Question 05c43

Feb 3, 2016

84.86%#

#### Explanation:

$2 F e {\left(H C {O}_{3}\right)}_{3} \rightarrow F {e}_{2} {\left(C {O}_{3}\right)}_{3} + 2 {H}_{2} O + 3 C {O}_{2}$

I will assume that the ${H}_{2} O$ and $C {O}_{2}$ are driven off as gases.

So the mass of ${H}_{2} O$ + $C {O}_{2} = 1.394 - 0.978 = 0.416 \text{g}$

We can work back and find how much iron(III) hydrogen carbonate would be needed to produce this amount of ${H}_{2} O$ and $C {O}_{2}$:

2 mol $F e {\left(H C {O}_{3}\right)}_{3} \rightarrow$ 2mol ${H}_{2} O$ + 3mol $C {O}_{2}$

Converting to grams using the ${M}_{r}$ values $\Rightarrow$

$\left(2 \times 238.9\right) \text{g"rarr(2xx18.01)"g"+(3xx44.00)"g}$

$\therefore 477.8 \text{g"rarr168.02"g}$

So this tells us that:

$168.02 \text{g}$ of $C {O}_{2}$ and ${H}_{2} O$ are produced from $477.8 \text{g}$ iron(III) hydrogen carbonate.

$\therefore 1 \text{g}$ from $\frac{477.8}{168.02} \text{g}$

$\therefore 0.416 \text{g}$ from $\frac{477.8}{168.02} \times 0.416 = 1.183 \text{g}$

This is the mass of iron(III) hydrogen carbonate.

So the % purity is:

$\frac{1.183}{1.394} \times 100 = 84.86$