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BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
nBaCl2=\(\dfrac{100.20,8\%}{208}=0,1\left(mol\right)\)
nNa2SO4=\(\dfrac{200.14,2\%}{142}=0,2\left(mol\right)\)
Vì 0,1<0,2 nên Na2SO4 dư 0,1 mol \(\Leftrightarrow\)14,2(g)
Theo PTHH ta có:
nBaCl2=nBaSO4=0,1(mol)
nNaCl=2nBaCl2=0,2(mol)
mBaSO4=233.0,1=23,3(g)
mNaCl=58,5.0,2=11,7(g)
mdd=100+200-23,3=276,7(g)
C% dd NaCl=\(\dfrac{11,7}{276,7}.100\%=4,23\%\)
C% dd Na2SO4=\(\dfrac{14,2}{276,7}.100\%=5,132\%\)
a)\(\text{PTHH: CaO + H}_2O->Ca\left(OH\right)_2\)
\(PTHH:Ca\left(OH\right)_2+CO_2->CaCO_3\downarrow+H_2O\)
\(PTHH:Ca\left(HCO_3\right)_2->CaCO_3+CO_2\uparrow+H_2O\)
b) \(n_{CâO}=\frac{3,92}{56}=0,07\left(mol\right)\)
\(n_{CO2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(\text{PTHH: CaO + H}_2O->Ca\left(OH\right)_2\)
0,07 mol 0,07 mol
Lập tỉ lệ : \(\frac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=\frac{0,1}{0,07}=1,43\)
=> Tạo muối \(Ca\left(HCO_3\right)_2;CaCO_3\)
Đặt \(n_{Ca\left(HCO_3\right)2}=a\left(mol\right);n_{CaCO_3}=b\left(mol\right)\)
\(PTHH:Ca\left(OH\right)_2+CO_2->CaCO_3\downarrow+H_2O\)
b mol b mol b mol
\(PTHH:Ca\left(OH\right)_2+2CO_2->Ca\left(HCO_3\right)_2\)
a mol 2a mol a mol
Theo đề ta có a + b = 0,07
2a + b = 0,1
=> a= 0,03 ; b=0,04
\(m_1=0,04\cdot100=4\left(g\right)\)
\(PTHH:Ca\left(HCO_3\right)_2->CaCO_3+CO_2\uparrow+H_2O\)
0,03 mol 0,03 mol
\(m_2=0,03\cdot100=3\left(g\right)\)
a) 2AgNO3+CaCl2---->2AgCl+Ca(NO3)2
n AgNO3=1,7/170=0,01(mol)
n CaCl2=2,22/111=0,02(mol)
----> CaCl2 dư
Theo pthh
n AgCl=n AgNO3=0,01(mol)
m AgCl=0,01.143,5=14,35(g)
V dd sau pư=70+30=`100ml=0,1(l)
n CaCl2 dư=0,02-0,005=0,015(mol)
CM CaCl2=0,015/0,1=0,15(M)
Theo pthh
n Ca(NO3)2=1/2 n AgCl=0,005(mol)
CM Ca(NO3)2=0,005/0,1=0,05(M)
Bài 2
BaCl2+H2SO4--->BaSO4+2HCl
a) n BaCl2=400.5,2/100=20,8(g)
n BaCl2=20,8/208=0,1(mol)
m H2SO4=100.1,14.20/100=22,8(g)
n H2SO4=22,8/98=0,232(mol)
---->H2SO4 dư
Theo pthh
n BaSO4=n BaCl2=0,1(mol)
m BaSO4=0,1.233=23,3(g)
b) m dd sau pư=400+114-23,3
=490,7(g)
Theo pthh
n HCl=2n BaCl2=0,2(mol)
C%HCl=\(\frac{0,2.36,5}{490,7}.100\%=1,88\%\)
n H2SO4 dư=0,232-0,1=0,132(mol)
C% H2SO4=\(\frac{0,132.98}{490,7}.100\%=2,64\%\)
B1:
\(n_{AgNO3}=0,01\left(mol\right);n_{CaCl2}=0,2\left(mol\right)\)
PTHH:\(2AgNO3+CaCl2\rightarrow2AgCl2\downarrow+Ca\left(NO3\right)2\)
Trước :0,01................0,02..........................................................(mol)
Pứng:\(0,01\rightarrow0,005\rightarrow0,01\rightarrow0,005\)
Dư: 0............................0,015......................................................(mol)
\(m\downarrow_{AgCL}=0,01.143,5=1,435\left(g\right)\)
Trong dd sau phản ứng chứa: \(\left\{{}\begin{matrix}Ca\left(NO3\right)2:0,005\left(mol\right)\\CaCl2:0,015\left(mol\right)\end{matrix}\right.\)
\(C_{M_{Ca\left(NO3\right)2}}=\frac{0,005}{0,1}=0,05M\)
\(C_{M_{CaCl2}}=\frac{0,015}{0,1}=0,15M\)
Bài 2:\(n_{BaCl2}=\frac{400.5,2}{100.208}=0,1\left(mol\right)\)
\(D=\frac{m_{dd}}{v_{dd}};C\%=\frac{m_{ct}}{m_{dd}}.100\Rightarrow m_{H2SO4}=\frac{D.v.d^2.C\%}{100}=22,8g\)
\(\Rightarrow n_{H2SO4}=0,23\left(mol\right)\)
\(BaCl2+HSO4\rightarrow BaSO4\downarrow+2HCl\)
0,1..............0,1............0,1.................0,2.....(mol)
\(a,m_{\downarrow}=0,1.223=23,3\left(g\right)\)
\(b,m_{dd_{saupu}}=m_{BaCl2}+m_{dd_{H2SO4}}-m_{\downarrow}_{BaSO4}\)
\(=400+1,14.100-23,3=490,7\)
\(\Rightarrow C\%_{HCl}=\frac{0,2.36,5}{490,7}.100\%=1,48\%\)
\(\%H2SO4_{du}=\frac{\left(0,23-0,1\right).98}{490,7}.100=2,59\%\)
a) CO2 +Ba(OH)2---->BaCO3 +H2O
b)n CO2 =0,1
nCO2 = nBa(OH)2 =0,1
----->Cm =0,5M
c)nCO2 = nBa(OH)2 =0,1
--->mBa(OH)2 =17,1
a ) \(CO_2+Ba\left(OH\right)_2--->BaCO_3+H_2O\)
b ) \(n_{CO_2}=0,1\)
\(n_{CO_2}=n_{Ba}\left(OH\right)_2=0,1\)
\(--->Cm=0,5M\)
c ) \(n_{CO_2}=n_{Ba}\left(OH\right)_2=0,1\)
\(--->m_{Ba}\left(OH\right)_2=17,1\).
Ta có\(\frac{m_{BaCl_2}}{100}.100\%=5,2\%\)
=> \(m_{BaCl_2}=5,2\left(g\right)\)
Lại có : \(\frac{m_{H_2SO_4}}{29,2}.100\%=20\%\Rightarrow m_{H_2SO_4}=5,84\left(g\right)\)
Phương trình hóa học phản ứng :
H2SO4 + BaCl2 ----> 2HCl + BaSO4
Tỉ lệ \(\frac{5,2}{1}< \frac{5,84}{1}\)
=> H2SO4 dư
\(n_{BaCl_2}=\frac{m}{M}=\frac{5,2}{208}=0,025\left(\text{mol}\right)\)
=> \(n_{BaSO_4}=0,025\left(mol\right)\)
=> Khối lượng chất kết tủa là : \(m_{BaSO_4}=n.M=0,025.233=5,825\left(g\right)\)
a) PTHH:\(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)
b) Ta có: \(n_{CaCO_3}=\frac{60}{100}=0,6\left(mol\right)\) \(\Rightarrow n_{CH_3COOH}=1,2mol\)
\(\Rightarrow m_{CH_3COOH}=1,2\cdot60=72\left(g\right)\) \(\Rightarrow m_{ddCH_3COOH}=\frac{72}{12\%}=600\left(g\right)\)
c) Theo PTHH: \(n_{CaCO_3}=n_{\left(CH_3COO\right)_2Ca}=n_{CO_2}=0,6mol\)
\(\Rightarrow\left\{{}\begin{matrix}m_{\left(CH_3COO\right)_2Ca}=0,6\cdot158=94,8\left(g\right)\\m_{CO_2}=0,6\cdot44=26,4\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{CaCO_3}+m_{ddCH_3COOH}-m_{CO_2}=633,6\left(g\right)\)
\(\Rightarrow C\%_{dd\left(CH_3COO\right)_2Ca}=\frac{94,8}{633,6}\cdot100\approx14,96\%\)
d) Ta có: \(n_{CO_2}=n_{BaCO_3}=0,6mol\)
\(\Rightarrow m_{BaCO_3}=0,6\cdot197=118,2\left(g\right)\)
PTHH 3ZnCl2+2H3PO4----->Zn3(PO4)2+6HCl
\(n_{ZnCl_2}\)=0,3.2=0,6(mol)
Theo phương trình =>\(\dfrac{1}{3}n_{ZnCl_2}=n_{Zn_3\left(PO_4\right)_2}=0,2\left(mol\right)\)
=>\(m_{Zn_3\left(PO_4\right)_2}\)=0,2.385=77(g)
Theo phương trình =>\(2n_{ZnCl_2}=n_{HCl}=1,2\left(mol\right)\)
=>\(C_{M_{HCl}}\)=\(\dfrac{1,2}{0,2+0,3}=2,4M\)