Bài 3 :
\(n_{H^+}=0.4\cdot1+0.4\cdot0.5\cdot2=0.8\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
\(0.8........0.8\)
\(n_{Ba\left(OH\right)_2}=\dfrac{0.8}{2}=0.4\left(mol\right)\)
\(V_{dd_{Ba\left(OH\right)_2}}=\dfrac{0.4}{2}=0.2\left(l\right)=200\left(ml\right)\)
\(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0.4\left(mol\right)\)
\(m_{BaSO_4}=0.4\cdot233=93.2\left(g\right)\)
Bài 2 :
\(n_{Fe_2O_3}=\dfrac{4.8}{160}=0.03\left(mol\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(0.03..........0.18......0.06\)
\(C_{M_{HCl}}=\dfrac{0.18}{0.6}=0.3\left(M\right)\)
\(m_{FeCl_3}=0.06\cdot162.5=9.75\left(g\right)\)
