Question

13.8 g of $K_{2}C{O}_3$ was treated by a series of reagents so as to covert all of its carbon to $K_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$. Calculate the mass (in g) of the product formed?
(Molecular mass of $K_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$ = 698 u)

Answer 1

Given,
$K_{2}C{O}_3\stackrel{severalsteps}{\to }{K}_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$

Since carbon atoms are conserved, applying Principle of Atom Conservation (POAC) for C atoms we get,
$\text{moles of C in}{K}_{2}C{O}_3=\text{moles of C in}{K}_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$

$\therefore 1\times \text{moles of}{K}_{2}C{O}_3=12\times \text{moles of}{K}_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$
$\Rightarrow \frac{\text{mass of}{K}_{2}C{O}_3}{\text{molar mass of}{K}_{2}C{O}_3}=12\times \frac{\text{mass of}{K}_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2}{\text{Molar mass of}{K}_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2}$
$\Rightarrow \frac{13.8}{138}=12\times \frac{\text{mass of}{K}_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2}{698}$
$\Rightarrow \text{mass of}{K}_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2=\frac{13.8}{138}\times \frac{698}{12}=5.82g$