\(S_{FeBr_2.6H_2O}\left(20^oC\right)=115\left(g\right)\)
\(\Rightarrow m_{ddFeBr_2.6H_2O}\left(20^oC\right)=115+100=215\left(g\right)\)
Ở 20oC: 215g dd FeBr2.6H2O có 115g FeBr2.6H2O
516g dd FeBr2.6H2O có x(g) FeBr2.6H2O
\(\Rightarrow x=m_{FeBr_2.6H_2O}=\frac{516\times115}{215}=276\left(g\right)\)