ta có: nBaCl2= \(\dfrac{20,8}{208}\)= 0,1( mol)
PTPU
BaCl2+ H2SO4\(\rightarrow\) BaSO4\(\downarrow\)+ 2HCl
.0,1...........0,1...........0,1...........0,2..... mol
\(\Rightarrow\) mBaSO4= 0,1. 233= 23,3( g)
ta có: mdd sau pư= mBaCl2+ mdd H2SO4- mBaSO4
= 20,8+ 200- 23,3
= 197,5( g)
ta có: mHCl= 0,2. 36,5= 7,3( g)
\(\Rightarrow\) C%HCl= \(\dfrac{7,3}{197,5}\). 100%= 3,67%
a) BaCl2 + H2SO4 → BaSO4↓ + 2HCl
b) \(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)
Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,1\times233=23,3\left(g\right)\)
c) Theo PT: \(n_{HCl}=2n_{BaCl_2}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2\times36,5=7,3\left(g\right)\)
\(m_{dd}saupư=m_{BaCl_2}+m_{ddH_2SO_4}-m_{BaSO_4}=20,8+200-23,3=197,5\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{7,3}{197,5}\times100\%=3,696\%\)