Question

cho 0.1 mol BaO tan trong 326.7 gam nước. tính C% dd sau phản ứng

Answer 1

\(m_{BaO}=15,3\left(g\right)\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

0,1                          0,1

\(\Rightarrow n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Ba\left(OH\right)_2}=17,1\left(g\right)\)

\(C\%_{Ba\left(OH\right)_2}=\dfrac{17,1}{326,7+15,3}=5\%\)