Fill in the blanks.
a. The equivalent weight of `NaHCO_(3)` is ……..and of `SO_(2)` is ………
b. 2 mol of 50% pure `Ca(HCO_(3))_(2)` on heating forms 1 mol of `CO_(2)`. The % yield of `CO_(2)` is ……..
c. `5 g` of `K_(2) SO_(4)` was dissolved in `250 mL` of solution. The volume of this solution that should be used so that `1.2 g` of `BaSO_(4)` be precipitated fromk `BaCl_(2)` is ....... (molecular mass of `K_(2) SO_(4) = 174` and `BaSO_(4) = 233`)
d. The residue obtained on strongly heating `2.76 g Ag_(2) CO_(3)` is ........
`[Ag_(2)CO_(3) overset(Delta)rarr Ag + CO_(2) + O_(2)]`
Atomic mass of `Ag = 108`

Answer

A, B, C, D

Explanation

a. 84,32 b. 100% c. `44.8 mL` d. `2.16 g`
a. `Ew of NaHCO_(3) = (84)/(1)` (valnecy factor = 2)
`Ew of SO_(2) = (64)/(2) = 32` (Valency factor = 2)
b. `underset(2 mol)(2Ca(HCO_(3))_(2)) rarr CaCO_(3) + underset(1 mol)(CO_(2)) + H_(2) O`
The percent yield of `CO_(2)` is 100% whether `Ca(HCO_(3))_(2)` is 100% pure or 50% pure.
c. `underset(1 mol)(K_(2) SO_(4)) + BaCl_(2) rarr underset(1 mol)(BaSO_(4)) + 2 KCl`
`M_(K_(2)SO_(4)) = (5 xx 1000)/(147 xx 250) = (20)/(174)`
`M xx V_(L) (K_(2) SO_(4)) =` moles of `BaSO_(4)`
`(20)/(174) xx V_(L) = (1.2)/(233)`
`V_(L) = 0.0488 L = 44.8 mL`
d. `underset((Mw = 276 g))(Ag_(2) CO_(3)) rarr (2 xx 108 g)(2Ag ) + CO_(2) uarr + O_(2) uarr`
The residue is due to `Ag`.
`276 g of Ag_(2) CO_(3) = 2 xx 108 g Ag`
`2.76 g of Ag_(2) CO_(3) = (2 xx 108 xx 2.76)/(276) = 2.16 g`