1.

27.6 g of K_(2) CO_(3)was treated by a series of reagents so as to convert all of its carbon to K_(2) Zn_(3) [Fe (CN)_(6)]_(2)Calculate the weight of the product.

Answer»

Solution : `K_(2) CO_(3) underset("steps")overset("several")to K_(2) Zn_(3) [Fe (CN)_(6)]_(2)`
SinceC , atoms are CONSERVE, applying POAC for C atoms,
moles of CIN `K_(2) CO_(3)` = moles of C in `K_(2) Zn_(3) [ Fe (CN)_(6)]_(2)`
`1 xx` molesof `K_(2) CO_(3) = 12 xx ` moles of `K_(2) Zn_(3) [Fe(CN)_(6)]_(2)`
(`:. 1` mole of `K_(2) CO_(3)` contains1 mole of Cand 1 mole of `K_(2) Zn_(3) [ Fe(CN)_(6)]_(2)` contains12 moles of C)
`(wt . of K_(2) CO_(3))/( mol . wt. of K_(2) CO_(3)) = 12 xx ("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`
[mol . wt. of `K_(2) CO_(3)` = 138 and mol. wt. of `K_(2)Zn_(3)[Fe (CN)_(6)]_(2)=698]`


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