Question

Answer 1

a) \(m_{BaCl_2}=100.20,8\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

             0,1----->0,1-------->0,1---------->0,2

b) \(V_{ddH_2SO_4}=\dfrac{0,1}{0,2}=0,5\left(l\right)\)

c) \(m_{dd\text{ }sau\text{ }phản\text{ }ứng}=100+100-0,1.233=176,7\left(g\right)\)

=> \(C\%_{HCl}=\dfrac{0,2.36,5}{176,7}.100\%=4,13\%\)