PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_{4\downarrow}+2H_2O\)
a, Ta có: \(n_{BaSO_4}=\dfrac{23,3}{233}=0,1\left(mol\right)\)
Theo PT: \(n_{Ba\left(OH\right)_2}=n_{BaSO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{Ba\left(OH\right)_2}=0,1.171=17,1\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{NaOH}=\dfrac{16}{16+17,1}.100\%\approx48,34\%\\\%m_{Ba\left(OH\right)_2}\approx51,66\%\end{matrix}\right.\)
b, Ta có: \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}+n_{Ba\left(OH\right)_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5M\)
Bạn tham khảo nhé!
