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![](https://rs.olm.vn/images/avt/0.png?1311)
\(C\%_{KCl}=\dfrac{20}{600}\cdot100\%=3.33\%\)
\(C\%_{K_2SO_4}=\dfrac{75}{1500}\cdot100\%=5\%\)
\(C\%_{NaCl}=\dfrac{15}{15+45}\cdot100\%=25\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) nNaCl = 1.0,5 = 0,5 (mol) → mNaCl = 0,5.(23 +35,5) = 29,25 (g)
b) nKNO3 = 2.0,5 = 1 (mol) → mKNO3 = 1.101 = 101 (g)
c) nCaCl2 = 0,1.0,25 = 0,025 (mol) → mCaCl2 = 0,025(40 + 71) = 2,775 (g)
d) nNa2SO4 = 0,3.2 = 0,6 (mol) → mNa2SO4 = 0,6.142 = 85,2 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)b)c)d) mBaCl2=150.16,64%=24,96g
=>nBaCl2=0,12 mol
mH2SO4=100.14,7%=14,7g=>nH2SO4=0,15mol
BaCl2 + H2SO4 =>BaSO4 +2HCl
Bđ: 0,12 mol; 0,15 mol
Pứ: 0,12 mol=>0,12 mol=>0,12 mol=>0,24 mol
Dư: 0,03 mol
Dd ban đầu chứa BaCl2 0,12 mol và H2SO4 0,15 mol
Dd A sau phản ứng chứa HCl 0,24 mol và H2SO4 dư 0,03 mol
mHCl=0,24.36,5=8,76g
mH2SO4=0,03.98=2,94g
Kết tủa B là BaSO4 0,12 mol=>mBaSO4=0,12.233=27,96g
mddA=mddBaCl2+mddH2SO4-mBaSO4
=150+100-27,96=222,04g
C%dd HCl=8,76/222,04.100%=3,945%
C% dd H2SO4=2,94/222,04.100%=1,324%
e) HCl +NaOH =>NaCl +H2O
0,24 mol=>0,24 mol
H2SO4 +2NaOH =>Na2SO4 + 2H2O
0,03 mol=>0,06 mol
TÔNG nNaOH=0,3 mol
=>V dd NaOH=0,3/2=0,15 lit
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a) \(C\%=\dfrac{15}{15+45}.100\%=25\%\)
b) \(C_M=\dfrac{0,5}{1,5}=0,33M\)
Câu 2:
a) \(n_{NaOH}=0,5.1=0,5\left(mol\right)=>m_{NaOH}=0,5.40=20\left(g\right)\)
b) \(n_{HCl}=0,2.0,5=0,1\left(mol\right)=>m_{HCl}=0,1.36,5=3,65\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(n_{KNO_3}=\dfrac{20}{101}=0,198\left(mol\right)\)
\(C_M=\dfrac{n}{V}=\dfrac{0,198}{0,85}=0,233M\)
Bài 2:
\(C_M=\dfrac{n}{V}=\dfrac{0,5}{0,75}=0,66M\)
Bài 3:
\(n_{KNO_3}=2.0,5=1\left(mol\right)\)
\(m_{KNO_3}=1.101=101\left(g\right)\)
Bài 4:
\(C\%=\dfrac{20}{600}.100=3,33\%\)
Bài 1:
\(n_{KNO_3}=\dfrac{20}{101}=0,198\left(mol\right)\)
\(C_{M_{ddKNO_3}}=\dfrac{0,198}{0,85}\approx0,23M\)
Bài 2:
\(C_{M_{ddKCl}}=\dfrac{0,5}{0,75}\approx0,667M\)
Bài 3:
\(n_{KNO_3}=0,5.2=1\left(mol\right)\Rightarrow m_{KNO_3}=1.101=101\left(g\right)\)
Bài 4:
\(C\%_{ddKCl}=\dfrac{20.100\%}{600}=3,333\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(C\%_{dd_{HCl}}=\dfrac{m_{HCl}}{mdd_{HCl}}\cdot100\%=\dfrac{20}{600}\cdot100\%=3,33\%\)
b)
\(mdd_{NaCl}=m_{NaCl}+m_{H_2O}=15+45=60\left(g\right)\)
=> \(C\%dd_{NaCl}=\dfrac{m_{NaCl}}{mdd_{NaCl}}\cdot100\%=\dfrac{15}{60}\cdot100\%=25\%\)
hok tốt nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\\a, ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ b,Vì:\dfrac{0,1}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\\ \Rightarrow m_{H_2SO_4\left(dư\right)}=98.0,3=29,4\left(g\right)\\ c,n_{ZnSO_4}=0,1.161=16,1\left(g\right)\\ m_{ddsau}=m_{ZnO}+m_{ddH_2SO_4}=8,1+200=208,1\left(g\right)\\ \Rightarrow C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{29,4}{208,1}.100\approx14,128\%\\ C\%_{ddZnSO_4}=\dfrac{16,1}{208,1}.100\approx7,737\%\)
ZnO+H2SO4->ZnSO4+H2O
0,1-----0,1-------0,1-------0,1 mol
n ZnO=\(\dfrac{8,1}{81}\)=0,1 mol
m H2SO4 =39,2g =>n H2SO4=\(\dfrac{39,2}{98}\)=0,4 mol
=>H2SO4 , dư 0,3 mol
=>m H2SO4=0,3.98=29,4g
=>C%H2SO4 dư=\(\dfrac{29,4}{200+0,1.18}\).100=14,568%
=>C% ZnSO4=\(\dfrac{0,1.161}{200+0,1.18}.100=7,9781\%\)
1.
a)
\(n_{KNO_3}=2.0,5=1\left(mol\right)\) => \(m_{KNO_3}=1.101=101\left(g\right)\)
b)
\(n_{CaCl_2}=0,25.0,1=0,025\left(mol\right)\)
=> \(m_{CaCl_2}=0,025.111=2,775\left(g\right)\)
2.
Nước tác dụng vs oxit axit --> dd axit
Nước tác dụng vs oxit bazơ-->dd bazơ
Nước tác dụng vs kim loại -->dd bazơ+hidro (những kloại có thể tác dụng Na, Ca , Ba ,K)
Bn tự vt PTHH đc ko ?
\(SO_2+H_2O-->H_2SO_3\)
\(Na_2O+H_2O-->2NaOH\)
\(Na+2H_2O-->NaOH+H_2\)
3.
a) \(C\%_{KCl}=\dfrac{20}{600}.100=3,34\%\)
b) \(C\%_{NaCl}=\dfrac{15}{45}.100=33,34\%\)