The amount of `BaSO_(4)` precipitated on mixing `BaCl_(2)` (0.5 M) with `H_(2)SO_(4)` (1M) will correspond to -

Options

A. 0.5 mol
B. 1.0 mol
C. 1.5 mol
D. 2.0 mol

Answer

A

Explanation

`[Ba^(2+)]=0.5mol*L^(-1),[SO_(4)^(2-)]=1mol*L^(-1)`
`Ba^(2+)(aq)+SO_(4)^(2-)(aq)hArr BaSO_(4)(s)`
0.5 mol of `Ba^(2+)` would react with 0.5 mol of `SO_(4)^(2-)` ions to form 0.5 mol of `BaSO_(4)`.