Y$=40$

In an experiment, $(120+40)mL0\mathrm{.}6MN{a}_{2}S{O}_4$ solution is mixed with $300mL0\mathrm{.}45MBaC{l}_2$ solution and $BaS{O}_4$ precipitate $($solid$)$ is formed according to the below reaction. Calculate the amount of precipitate in grams. $($Atomic mass: $Na=23,O=16,S=32,Ba=137,Cl=35)$

$N{a}_{2}S{O}_4(aq)+BaC{l}_2(aq)2$NaCl$(aq)+BaS{O}_4(s)$