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a)
$BaCl_2 + H_2SO_4 \to BaSO_4 + 2HCl$
$n_{BaCl_2} = 0,1 < n_{H_2SO_4} = 0,2$ nên $H_2SO_4$ dư
$n_{BaSO_4} = n_{BaCl_2} = 0,1(mol)$
$m_{BaSO_4} = 0,1.233 = 23,3(gam)$
b)
A gồm :
$HCl : 0,1.2 = 0,2(mol)$
$H_2SO_4\ dư : 0,2 - 0,1 = 0,1(mol)$
$V_{dd} = 0,1 + 0,1= 0,2(lít)$
$C_{M_{HCl}} = \dfrac{0,2}{0,2} = 1M$
$C_{M_{H_2SO_4}} = \dfrac{0,1}{0,2} = 0,5M$
c)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4\ dư} = 0,2(mol)$
$m_{dd\ NaOH} = \dfrac{0,2.40}{15\%} = 53,33(gam)$
mddH2SO4 = 100 . 1,137 = 113,7
nH2SO4 = 113,7 . 20%/98 = 0,232 mol
nBaCl2 = 400 . 5,29%/208 = 0,1 mol
H2SO4 + BaCl2 —> BaSO4 + 2HCI
Bđ: 0,232 0,1
Pứ: 0,1 0, 1 0,1 0,2
Sau pứ: 0,132 0
mBaSO4 = 0,1.233 = 23,3 gam
Khối lượng dung dịch sau khi lọc bỏ kết tủa:
mddB = mddH2SO4 + mddBaCl2 - mBaSO4 = 490,4
C%HCI = 0,2.36,5/490,4 = 1,49%
C%H2SO4 dư = 0,132.98/490,4 = 2,64%
\(n_{BaCl_2}=\dfrac{208.10\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ 0,1............0,1..............0,1.............0,2\left(mol\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{8}=122,5\left(g\right)\\ c,m_{kt}=m_{BaSO_4}=0,1.233=23,3\left(g\right)\\ d,m_{ddsau}=208+122,5-23,3=307,2\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{307,2}.100\approx2,376\%\)
Na2SO4 + BaCl2 →2NaCl + BaSO4
nNa2SO4=0,05.0,1=0,005(mol)
nBaCl2=0,1.0,1=0,01(mol)
Vì 0,005<0,01 nên BaCl2 dư 0,005(mol)
Theo PTHH ta có;
nNa2SO4=nBaSO4=0,005(mol)
2nNa2SO4=nNaCl=0,01(mol)
mBaSO4=0,005.233=1,165(g)
CM dd BaCl2=\(\dfrac{0,005}{0,15}=\dfrac{1}{30}\)M
CM dd NaCl=\(\dfrac{0,01}{0,15}=115\)M
\(a)n_{BaCl_2}=\dfrac{240}{1,12}:1000\cdot1=\dfrac{3}{14}mol\\ n_{H_2SO_4}=\dfrac{122.20}{100}:98=\dfrac{61}{245}mol\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ \Rightarrow\dfrac{3:14}{1}< \dfrac{61:245}{1}\Rightarrow H_2SO_4.dư\\ n_{BaSO_4}=n_{BaCl_2}=n_{H_2SO_4}=\dfrac{3}{14}mol\\ m_{kt}=m_{BaSO_4}=\dfrac{3}{14}\cdot233=50g\\ c)C_{\%H_2SO_4\left(dư\right)}=\dfrac{\left(61:245-3:14\right)98}{240+122-50}\cdot100=1,2\%\)
\(m_{H_2SO_4}=\dfrac{19,6\cdot20\%}{100\%}=3,92\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\\ PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \Rightarrow n_{H_2SO_4}=n_{BaCl_2}=n_{BaSO_4}=0,04\left(mol\right)\\ \Rightarrow m_{CT_{BaCl_2}}=0,04\cdot208=8,32\left(g\right)\\ \Rightarrow m_{dd_{BaCl_2}}=\dfrac{8,32\cdot100\%}{12\%}\approx69,3\left(g\right)\\ m_{kết.tủa}=m_{BaSO_4}=0,04\cdot233=9,32\left(g\right)\)
\(n_{BaCl_2}=\dfrac{150.16,64\%}{137+35.2}=0,12\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.14,7\%}{98}=0,15\left(mol\right)\)
Phương trình hóa học :
BaCl2 + H2SO4 -----> BaSO4 + 2HCl
Dễ thấy \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{1}\Rightarrow H_2SO_4\text{ dư }0,15-0,12=0,03\left(mol\right)\)
c) Khối lượng kết tủa :
\(m_{BaSO_4}=0,12.233=27,96\) (g)
Khối lượng chất tan : \(m_{HCl}=0,24.36,5=8,76\left(g\right)\) ;
\(m_{H_2SO_4\left(\text{dư}\right)}=0,03.98=2,94\left(g\right)\)
c) \(C\%_{H_2SO_4}\)= \(\dfrac{2,94}{150+100}.100\%=1,176\%\)
\(C\%_{HCl}=\dfrac{8,76}{150+100}.100\%=3.504\%\)
d) NaOH + HCl ---> NaCl + H2O
0,24 <-- 0,24
mol mol
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
0,06 mol <-- 0,03 mol
\(\Rightarrow n_{NaOH}=0,24+0,06=0,3\left(mol\right)\)
\(V_{NaOH}=0,3.2=0,6\left(l\right)\)
nH2SO4=1*0,1=0,1(mol)
PTHH:
BaCl2 + H2SO4 => BaSO4 + 2HCL
1 1 1 1 (mol)
0,1 0,1 0,1 0,1 (mol)
=> kết tủa thu đc là BaSO4
a, Theo PTHH:
nBaSO4=0,1 (mol) => mBaSO4=0,1*233=23,3 (g)
b, Theo PTHH: nBaCl2= 0,1 (mol)
=> CM= 0,1/0,2 = 0,5M
1.
a, \(n_{H_2SO_4}=0,1.1=0,1\left(mol\right)\)
PTHH: H2SO4 + BaCl2 → BaSO4 + 2HCl
Mol: 0,1 0,1 0,1
\(m_{BaSO_4}=233.0,1=23,3\left(g\right)\)
b, \(C_{M_{ddBaCl_2}}=\dfrac{0,1}{0,2}=0,5M\)