30 mL of `0.2 N BaCl_(2)` is mixed with 40 mL of `0.3 N Al_(2)(SO_4)_(3)`. How many g of `BaSO_(4)` are formed?

Explanation

First balance it
`{:(,3BaCl_(2)+,Al_(2)(SO_(4))_(3)rarr,3BaSO_(4)+,2AlCl_(3)),(mM"before",30xx0.1,40xx0.2,,),("reaction",=3,=8,0,0),(mM"after reaction",0,(8-1),3,2):}`
`:' 3 mM "of" BaCl_(2)` reacts with one mole of `Al_(2)(SO_(4))_(3)`
`:. mM "of" BaSO_(4) "formed" =3=(wt.)/(mol.wt.)xx1000`,
`:. wt. "of" BaSO_(4)"formed"=(3xx233)/(1000)=0.6999 g`
The above question may also be solved in terms of normality. We have
Molarity `xx` Valence factor = Normality.
Also balancing of equation is not neccessary in this concept.
`{:(,BaCI_(2)+,AI_(2)(SO_(4))_(3)rarr,BaSO_(4)+,AICI_(3)),("Meq.before",30xx0.1xx2,40xx0.2xx6,,),("reaction",=6,=48,0,0),("Meq.left after",,,,),(reaction,0,42,6,6):}`
Meq. of `BaSO_(4)` formed `= 6 = (wt.)/(Eq.wt) xx 1000`
`w = (6 xx 233)/(2xx1000) = 0.699g`