Question

cho 11 g dung dich \(H_2SO_420\%\) tac dung voi 400g dung dich Ba\(Cl_25,2\%\)
tinh Ba\(SO_4\)?

tinh C% dung dich sau PU khong ke ket tua

Answer 1

\(n_{H_2SO_4}=\dfrac{11.20\%}{98}=0,02\left(mol\right)\)

\(n_{BaCl_2}=\dfrac{400.5,2\%}{208}=0,1\left(mol\right)\)

\(H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\)

0,02mol 0,1mol -----> 0,02 ---> 0,04

Lập tỉ số: \(n_{H_2SO_4}:n_{BaCl_2}=0,02< 0,1\)

=> H₂SO₄ hết, BaCl₂ dư

\(n_{BaCl_2\left(dư\right)}=0,1-0,02=0,08\left(mol\right)\)

\(m_{BaSO_4}=0,02.233=4,66\left(g\right)\)

\(C\%_{BaCl_2\left(dư\right)}=\dfrac{0,08.208.100}{400}=4,16\%\)

\(C\%_{HCl}=\dfrac{0,04.36,5.100}{11+400-4,66}=0,36\%\)