Câu 6:
Ta có: \(n_{HCl}=xV_1\left(mol\right)\)
\(n_{H_2SO_4}=0,2xV_2\left(mol\right)\)
Khi trộn 2 dd với nhau, ta có: \(C_{M_{HCl}}=\dfrac{xV_1}{V_1+V_2}\left(M\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0,2xV_2}{V_1+V_2}\left(M\right)\)
Mà: CM (HCl) = 3CM (H2SO4) \(\Rightarrow\dfrac{xV_1}{V_1+V_2}=3.\dfrac{0,2xV_2}{V_1+V_2}\)\(\Rightarrow\dfrac{V_1}{V_2}=\dfrac{3}{5}\)
Vậy: Phải trộn 2 dd theo tỉ lệ thể tích là 3:5.
a, Trong 200 ml dd B có 75 ml dd HCl và 125 ml dd H2SO4.
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,075x\left(mol\right)\\n_{H_2SO_4}=0,125.0,2x\left(mol\right)\end{matrix}\right.\)
Ta có: \(m_{ddBa\left(OH\right)_2}=166,67.1,2=200\left(g\right)\)
\(\Rightarrow n_{Ba\left(OH\right)_2}=\dfrac{200.17,1\%}{171}=0,2\left(mol\right)\)
PT: \(2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\)
\(H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_{4\downarrow}+2H_2O\)
Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}+n_{H_2SO_4}\) \(\Rightarrow0,075x.\dfrac{1}{2}+0,125.0,2x=0,2\)
⇒ x = 3,2
Vậy: CM (HCl) = 3,2 và CM (H2SO4) = 3,2.0,2 = 0,64 (M)
b, Ta có: \(n_{H_2SO_4\left(B\right)}=0,125.0,2.3,2=0,08\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4\left(B\right)}}=\dfrac{0,08}{0,2}=0,4\left(M\right)\)
