`27.6gK_(2)CO_(3)` was treated by a series of reagents so as to convent all of its carbon to `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)` Calculate the weight of the product [mol.wt. of `K_(2)CO_(3)=138` and mol. Wt. of `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)=698]` .

Explanation

Here we have no knowledge about series of chemical reactions but we know initial reactant and final product, accordingly.
`K_(2)CO_(3) underset("Steps")overset("Several")to K_(2)Zn_(3)[Fe(CN)_(6)]_(2)`
Since C atoms are conserved, applying POAC for C atoms,
moles of C in `K_(2)CO_(3)` =moles of C in `K_(2)Zn_(3) [Fe(CN)_(6)]_(2)`
`1 xx ` moles of `k_(2)CO_(3)=12 xx` moles of `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)`
`(because ` 1 mole of `K_(2)CO_(3)` contains 1 moles of `C)`
`("wt. of" K_(2)CO_(3))/("mol. wt. of" K_(2)CO_(3))=12xx("wt. of the product")/("mol. wt. of product")`
wt. of `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)=(27.6)/(138)xx(698)/(12)=11.6 g`