`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 konowledge about series of chemical reactions but we know about initiL reactant and final product accordingly
`K_(2)CO_(3)underset("Steps")overset("Several")rarrK_(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)`
`(therefore 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) = 11.6g` .