a) PTHH
\(H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\)(1)
b)\(m_{H_2SO_4}=\dfrac{14,7\times100}{100}=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
\(m_{BaCl_2}=\dfrac{20,8\times200}{100}=41,6\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{208}{1165}\left(mol\right)\)
So sánh ta thấy \(n_{BaCl_2\left(du\right)}=\dfrac{133}{4660}\left(mol\right)\)
Theo (1) \(n_{BaSO_4}=0,15\left(mol\right)\Rightarrow m_{BaSO_4}=34,95\left(g\right)\)
c) Ta có chất tan gồm \(BaCl_2\)(dư) và HCl
\(m_{BaCl_2}=\dfrac{133}{4660}\times208\approx5,94\left(g\right)\)
Theo (1)\(n_{HCl}=0,3\left(mol\right)\Rightarrow m_{HCl}=10,95\left(g\right)\)
\(m_{H_2SO_4}=100\times14,7\%=14,7\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
\(m_{BaCl_2}=200\times20,8\%=41,6\left(g\right)\)
\(\Rightarrow n_{BaCl_2}=\dfrac{41,6}{208}=0,2\left(mol\right)\)
a) PTHH: H2SO4 + BaCl2 → BaSO4↓ + 2HCl
Ban đầu: 0,15.............0,2.........................................(mol)
Phản ứng: 0,15.............0,15........................................(mol)
Sau phản ứng: 0................0,05...→...0,15.............0,3.....(mol)
b) \(m_{BaSO_4}=0,15\times233=34,95\left(g\right)\)
c) \(m_{HCl}=0,3\times36,5=10,95\left(g\right)\)
d) Dung dịch sau phản ứng gồm: BaCl2 dư và HCl
\(m_{dd}saupư=100+200-34,95=265,05\left(g\right)\)
\(m_{BaCl_2}dư=0,05\times208=10,4\left(g\right)\)
\(\Rightarrow C\%_{BaCl_2}dư=\dfrac{10,4}{265,05}\times100\%=3,92\%\)
\(C\%_{HCl}=\dfrac{10,95}{265,05}\times100\%=4,13\%\)
