50 mL solution of `BaCl_(2)` (20.8% w//v) and 100 mL solution of `H_(2)SO_(4)` (9.8% w//v) are mixed (Ba = 137, Cl = 35.5, S=32)
`BaCl_(2) + H_(2)SO_(4) rightarrow BaSO_(4) darr 2HCl`
Weight of `BaSO_(4)` formed is:

Options

A. 11.65 g
B. 23.3 g
C. 29.8 g
D. 46.6 g

Answer

A

Explanation

Molecular mass of `BaCl_(2)=137+2xx35.5=208`
Amount of `BaCl_(2)` present in 50 mL solution
`=(20.8)/(2)=10.4g=(10.4)/(208)"mole"="0.05 mole"`
Amount of `H_(2)SO_(4)` present in 100 mL solution
`=9.8g=(9.8)/(98)"mole"="0.1 mole"`
`BaCl_(2)+H_(2)SO_(4)rarrBaSO_(4)+2HCl`
Thus, `BaCl_(2)` will be limiting reagent.
0.05 mole of `BaCl_(2)` will be produce `BaSO_(4)`
= 0.05 mole
`=0.05xx233 g` (mol. mass of `BaSO_(4)=137+32+64=233`)
= 11.65 g