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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
nNa = 6.9 : 23 = 0.3 mol
4Na + O2 ->2 Na2O
mol : 0.3 -> 0.15
Na2O + H2O -> 2NaOH
mol : 0.15 -> 0.3
mdd = 0.15 x 62 + 140.7 = 150g
C% NaOH = 0.3x40: 150 x 100% = 8%
\(n_{Na_2SO_4}=\dfrac{10\%.335}{142}\) số lẻ ơ
Bài 19 :
\(a) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)\\ V_{H_2} = 0,6.22,4 = 13,44(lít)\\ b) \text{Chất tan : }Al_2(SO_4)_3\\ n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol)\\ m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)\)
Bài 18 :
\(a) n_{HCl} = \dfrac{250.7,3\%}{36,5 } = 0,5(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol) \Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)\\ b) \text{Chất tan : } ZnCl_2\\ n_{ZnCl_2} = n_{H_2} = 0,25(mol)\\ m_{ZnCl_2} = 0,25.136 = 34(gam)\)
5. \(n_{BaCl_2}=\dfrac{400.5,2\%}{208}=0,1\left(mol\right)\)
\(m_{dd}=1,14.100=114\left(g\right)\)
\(m_{H_2SO_4}=114.20\%=22,8\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{22,8}{98}=0,23\left(mol\right)\)
\(Pt:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
0,1 mol 0,23mol \(\rightarrow0,1mol\) \(\rightarrow0,2mol\)
Lập tỉ số: \(n_{BaCl_2}\) : \(n_{H_2SO_4}=0,1< 0,23\)
\(\Rightarrow BaCl_2\) hết; \(H_2SO_4\) dư
\(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
\(n_{H_2SO_4\left(dư\right)}=0,23-0,1=0,13\left(mol\right)\)
\(\Sigma_{hh\left(spu\right)}=400+114-23,3=490,7\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,13.98.100}{490,7}=2,6\%\)
\(C\%_{HCl}=\dfrac{0,2.36,5.100}{490,7}=1,49\%\)
6. \(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)
\(n_{HCl}=2.0,2=0,4\left(mol\right)\)
Pt: \(NaOH+HCl\rightarrow NaCl+H_2O\)
0,5mol 0,4mol
Lập tỉ số: \(n_{NaOH}\) : \(n_{HCl}=0,5>0,4\)
\(\Rightarrow NaOH\) dư; HCl hết
\(n_{NaOH\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)
\(m_{NaOH\left(dư\right)}=0,1.40=4\left(g\right)\)
7. \(n_{Ba\left(OH\right)_2}=\dfrac{300.30\%}{171}=0,52\left(mol\right)\)
Pt: \(Ba\left(OH\right)_2+FeSO_4\rightarrow BaSO_4+Fe\left(OH\right)_2\)
0,52mol \(\rightarrow0,52mol\)\(\rightarrow0,52mol\)
\(m_{BaSO_4}=0,52.233=121,16\left(g\right)\)
\(m_{Fe\left(OH\right)_2}=0,52.90=46,8\left(g\right)\)
\(\Sigma_{hh\left(spu\right)}=300+800-121,16=978,84\left(g\right)\)
\(C\%_{Fe\left(OH\right)_2}=\dfrac{46,8.100}{978,84}=4,78\%\)
\(n_{BaCl_2}=\frac{400.5,2\%}{208}=0,1\left(mol\right);n_{H_2SO_4}=\frac{100.1,14.19,6\%}{98}=0,228\left(mol\right)\)
PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
Theo đề: 0,1.........0,228.....................................(mol)
Lập tỉ lệ: \(\frac{0,1}{1}< \frac{0,228}{1}\)=> Sau phản ứng H2SO4 dư
Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
dd sau khi lọc bỏ kết tủa: H2SO4 dư, HCl
\(m_{ddsaup.ứ}=400+114-23,3=490,7\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4\left(dư\right)}=\frac{\left(0,228-0,1\right).98}{490,7}.100=2,56\%\)
\(C\%_{HCl}=\frac{0,1.2.36,5}{490,7}.100=1,49\%\)
\(n_{CaCl_2}=0,1.1,5=0,15\left(mol\right)\)
\(PTHH:Na_2CO_3+CaCl_2\rightarrow2NaCl+CaCO_3\)
(mol)_____0,15_________0,15_____0,3_______0,15
\(m_{\downarrow\left(CaCO_3\right)}=100.0,15=15\left(g\right)\)
\(C_{M_{Na_2CO_3}}=\frac{0,15}{0,05}=3\left(M\right)\)
\(C_{M_B}=\frac{0,3}{0,15}=2\left(M\right)\)
mNa2CO3=100*19.96/100=19.96g
nNa2CO3=19.96/106=0.19mol
mBaCl2=200*10.04/100=20.08g
nBaCl2=20.08/208=0.097mol
Na2CO3 + BaCl2 -> BaCO3 + 2NaCl
(mol) 1 1
(mol) 0.19 0.097
Lập tỉ lệ: 0.19> 0.097. Na2Co3 dư dư
Na2CO3 + BaCl2 -> BaCO3 + 2NaCl
(mol) 0.097 0.097 0.097 0.194
mdd = mddNa2CO3 + mddBaCl2 - mBaCO3
=100+200-0.097*197=208.891g
nNa2CO3 dư = 0.19-0.097=0.093mol
mNa2CO3 dư = 0.093*106=9.858g
mNaCl = 0.194*58.5=11.349g
C%Na2CO3 dư = 9.858/208.891*100=4.72%
C%NaCl = 11.349/208.891*100=5.43%
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