For the reaction,
`2Na_(2)PO_(4(aq))+3Ba(PO_(3))_(2(aq))rarrBa_(3)(PO_(4))_(2_(s))+6NaNO_(3(aq))`
Suppose that a solution contaning `32.8 g of Na_(3)PO_(4)` and `26.1 g of Ba(NO_(3))_(2)` is mixed. How many `g` of `Ba_(3)(PO_(4))_(2)` are formed?

Explanation

`{:(,2Na_(3)PO+,3Ba(NO_(3))_(2)rarr,BA_(3)(PO)_(4)+,6NaNO_(3)),("Mole",(32.8^(*)//164),26.1//261,0,0),("before",,,,),("reaction",=0.2,0.1,0,0),("Mole",[0.2 - (0.1 xx2)//3],0,0.1//3,(0.1xx6)//3):}`
after reaction
Mole ration for
`Na_(3)PO_(4):Ba(NO_(3))_(2):Ba_(3)(PO_(4))_(2):NaNO_(3)::2:3:1:6`
Mole of `Ba_(3)(PO_(4))` formed `= 0.1//3`
`:.` Weight of `Ba_(3)(PO_(4))_(2)` formed `=(0.1)/(3)xx602`
`20.07 g`