vẽ hình ra đi bn k thì ghi cái nào nt hay // cg dc

\(n_{BaCl_2}=0,12\left(mol\right)\)

\(n_{H_2SO_4}\approx0,1438\left(mol\right)\)

\(n_{Na_2CO_3}=0,1\left(mol\right)\)

\(BaCl_2+H_2SO_4-->BaSO_4+2HCl\)

\(Na_2CO_3+H_2SO_4-->Na_2SO_4+CO_2+H_2O\)

\(\dfrac{0,1}{1}>\dfrac{0,1438-0,12}{1}\) => Na2CO3 dư

\(m_{ddB}=10,6+100-\left(0,1438-0,12\right)44=109,5528\left(g\right)\)

\(m_{ddA}=150+100-0,12.233=222,04\left(g\right)\)

\(\%H_2SO_4du=\dfrac{98.0,0238}{222,04}.100\%\approx1,05\%\)

\(\%HCl=\dfrac{0,12.36,5}{222,04}.100\%\approx1,97\%\)

b) \(HCl+NaOH-->NaCl+H_2O\)

0,12.............0,12

\(H_2SO_4+2NaOH-->Na_2SO_4+2H_2O\)

0,0238.......0,0476

\(V_{NAOH}=\dfrac{0,12+0,0476}{2}=0,0838\left(ml\right)=83,8\left(l\right)\)