Ta có:
\(\left\{{}\begin{matrix}n_{Ba\left(NO3\right)2}=\frac{26,1}{261}=0,1\left(mol\right)\\n_{Na2SO4}=\frac{28,4}{142}=0,2\left(mol\right)\end{matrix}\right.\)
\(PTHH:Ba\left(NO_3\right)_2+Na_2SO_4\rightarrow2NaNO_3+BaSO_4\)
Nên Na2SO4 dư
\(\Rightarrow n_{BaSO4}=n_{Ba\left(NO3\right)2}=0,1\left(mol\right)\)
\(\Rightarrow m=m_{BaSO4}=0,1.233=23,3\left(g\right)\)