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 $K_{2} Z n_{3} [F e (C N)_{6}]_{2}$ . Calculate the mass (in g) of the product formed?
(Molecular mass of $K_{2} Z n_{3} [F e (C N)_{6}]_{2}$ = 698 u)

Open in App

Solution

Given,
$K_{2} C O_{3} s e v e r a l s t e p s - ------ \to K_{2} Z n_{3} [F e (C N)_{6}]_{2}$

Since carbon atoms are conserved, applying Principle of Atom Conservation (POAC) for C atoms we get,
$moles of C in K_{2} C O_{3} = moles of C in K_{2} Z n_{3} [F e (C N)_{6}]_{2}$

$∴ 1 \times moles of K_{2} C O_{3} = 12 \times moles of K_{2} Z n_{3} [F e (C N)_{6}]_{2}$
$\Rightarrow \frac{mass of K_{2} C O_{3}}{molar mass of K_{2} C O_{3}} = 12 \times \frac{mass of K_{2} Z n_{3} [F e (C N)_{6}]_{2}}{Molar mass of K_{2} Z n_{3} [F e (C N)_{6}]_{2}}$
$\Rightarrow \frac{13.8}{138} = 12 \times \frac{mass of K_{2} Z n_{3} [F e (C N)_{6}]_{2}}{698}$
$\Rightarrow mass of K_{2} Z n_{3} [F e (C N)_{6}]_{2} = \frac{13.8}{138} \times \frac{698}{12} = 5.82 g$

Suggest Corrections

Similar questions

K_{2} C O_{3}

K_{2} Z n_{3} [F e (C N)_{6}]_{2}

K_{2} Z n_{3} [F e (C N)_{6}]_{2}

27.6

K_{2} C O_{3}

K_{2} {Z n}_{3} {[F e {(C N)}_{6}]}_{2}

K_{2} C O_{3}

K_{3} Z n_{3} [F e (C N)_{6}]_{2}

K_{2} C O_{3}

K_{3} Z n_{3} [F e (C N)_{6}]_{2}

3 Z n^{2 +} + 2 K_{4} [F e (C N)_{6}] \to K_{2} Z n_{3} [F e (C N)_{6}]_{2} + 6 k^{+}

K_{4} [F e (C N)_{6}]

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program