Question

24 g of $K_{2}C{O}_3$ was treated by a series of reagents to convert all of the carbon to cyanide ions in the complex $K_{3}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$. Calculate the weight of the complex formed.
(Molar mass of K = 39 g/mol, Fe = 56 g/mol, Zn = 65 g/mol)

• A. 15.85 g
• B. ​​​​​10.67 g
• C. 6.35 g
• D. 20.74 g

Answer 1

The correct option is B

​​​​​10.67 g

Let’s assume the weight of the complex compound to be $x$ g.
$K_{2}C{O}_3⟶K_{3}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$

According to the Principle of Atom Conservation (POAC) the number of carbon atoms is conserved.

Applying POAC on $C$,
$1\times \text{moles of}{K}_{2}C{O}_3=12\times \text{moles of}{K}_{3}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$

$1\times \frac{\text{weight of}{K}_{2}C{O}_3}{\text{molar mass of}{K}_{2}C{O}_3}=12\times \frac{\text{weight of the complex compound}}{\text{molar mass of the complex compound}}$

$1\times \frac{24}{138}=12\times \frac{x}{736}$

On solving, $x=10.67$ g
Hence, weight of the complex compound is 10.67 g.