To 100 " mL of " `KMnO_4` solution containing 0.632 g of `KMnO_4` 200 " mL of " `SnCl_2` containing 2.4g is added in presence of HCl. To the resulting solution, an excess of `HgCl_2` is added at once. How many grams of `Hg_2Cl_2` will be precipitated? (molarcular mass of `KMnO_4` is 158, `SnCl_2` is 95, and `Hg_2Cl_2` is 471 g `mol^(-1))`

Explanation

100 mL `KMnO_4+200mLSnCl_2+H^(o+)toMn^(2+)+Sn^(4+)`
m" Eq of "`KMnO_4=(0.632xx1000)/((158)/(5))=20`
`m" Eq of "SnCl_2=(2.4)/((190)/(2))xx1000=25.26`
`impliesSnCl_2` is in excess
`{:(Excess m" Eq of "SnCl_2=5.26),(m" mol of "SnCl_2=(1)/(2)xx5.26=2.63):}]`
`SnCl_2+2HgCl_2toHg_2Cl_2+SnCl_4`
`implies1molSnCl_2-=1 mol Hg_2Cl_2`
`impliesmolHg_2Cl_2 formed=2.6xx10^(-3)`
`-=2.6xx10^(-3)xx471g`
`=1.225g`