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

Answer

D

Explanation

`underset(1mol)(BaCl_(2))+underset(1mol)(H_(2)SO_(4))tounderset(1mol)(BaSO_(4))+2HCl`
Mass of `BaCl_(2) = (50xx20.8)/(100)g`
Moles of `BaCl_(2) = (50xx20.8)/(208xx100)=0.05 mol`
Mass of `H_(2)SO_(4) = (100xx9.8)/(100) = 9.8g`
Moles of `H_(2)SO_(4) = (9.8)/(98) = 0.1 mol`
Here `BaCl_(2)` is the limiting reactant
1 mol of `BaCl_(2)` gives 1 mol of `BaSO_(4)`
`:.` Mass of `BaSO_(4)`
`= 0.05 molxx[(137+32+64)gmol^(-1)]`
`0.05molxx233gmol^(-1) = 11.65g`