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Bài 3 :
\(n_{CaCO3}=\dfrac{10}{100}=0,1\left(mol\right)\)
Pt : \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O|\)
1 2 1 1 1
0,1 0,2 0,1 0,1
a) \(n_{CO2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,1.24,79=2,479\left(l\right)\)
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)
c) \(n_{CaCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{CaCl2}=0,1.111=11,1\left(g\right)\)
\(m_{ddspu}=10+100-\left(0,1.44\right)=105,6\left(g\right)\)
\(C_{CaCl2}=\dfrac{11,1.100}{105,6}=10,51\)0/0
Chúc bạn học tốt
Bài 11:
\(PTHH:2A+Cl_2\rightarrow2ACl\\TheoĐLBTKL:\\ m_A+m_{Cl_2}=m_{ACl}\\ \Leftrightarrow 9,2+m_{Cl_2}=23,4\\ \Rightarrow m_{Cl_2}=23,4-9,2=14,2\left(g\right)\\ n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\\ n_A=2.0,2=0,4\left(mol\right)\\ M_A=\dfrac{9,2}{0,4}=23\left(\dfrac{g}{mol}\right)\\ \Rightarrow A\left(I\right):Natri\left(Na=23\right)\)
a)
4Na + O2 ---to→ 2Na2O
Na2O + H2O → 2NaOH
2NaOH + CO2 → Na2CO3 + H2O
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
CaCO3 ---to→ CaO + CO2
CO2 + NaOH → NaHCO3
NaHCO3 + H2SO4 → Na2SO4 + CO2 + H2O
Na2SO4 + Ba(OH)2 → 2NaOH + BaSO4
b)
S + O2 ---to→ SO2
2SO2 + O2 ---to(V2O5)→ 2SO3
SO3 + H2O → H2SO4
H2SO4 + Cu(OH)2 → CuSO4 + 2H2O
CuSO4 + FeCl2 → CuCl2 + FeSO4
FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4
Fe(OH)2 + 2HCl → FeCl2 + 2H2O
2FeCl2 + Cl2 → 2FeCl3
2FeCl3 + 3Ba(OH)2 → 2Fe(OH)3 + 3BaCl2
2Fe(OH)3 ---to→ Fe2O3 + 3H2O
Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
43.a) \(m_{HCl\left(bđ\right)}=200.10,95\%=21,9\left(g\right)\)
=> \(n_{HCl\left(bđ\right)}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
b) HCl phản ứng với NaOH là HCl dư
\(HCl+NaOH\rightarrow NaCl+H_2O\)
\(n_{HCl\left(dư\right)}=n_{NaOH}=0,05.2=0,1\left(mol\right)\)
=> \(n_{HCl\left(pứ\right)}=n_{HCl\left(bđ\right)}-n_{HCl\left(dư\right)}=0,6-0,1=0,5\left(mol\right)\)
c) \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
\(n_{CaCO_3}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)
=> \(m_{CaCO_3}=0,25.100=25\left(g\right)\)
d) \(n_{CO_2}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)
=> \(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)
e) \(m_{ddsaupu}=25+200-0,25.44=214\left(g\right)\)
Dung dịch A gồm CaCl2 và HCl dư
\(n_{CaCl_2}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)
\(C\%_{CaCl_2}=\dfrac{0,25.111}{214}.100=12,97\%\)
\(C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{214}.100=1,71\%\)
Câu 7:
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\\\%m_{CuO}=44\%\end{matrix}\right.\)
c, \(n_{CuO}=\dfrac{10-0,1.56}{80}=0,055\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{CuO}=0,155\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,155.98}{100}.100\%=15,19\%\)
d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,055\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2\left(g\right)\\m_{CuSO_4}=0,055.160=8,8\left(g\right)\end{matrix}\right.\)
Câu 8:
a, \(CuCO_3+2HCl\rightarrow CuCl_2+CO_2+H_2O\)
b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{CuCO_3}=n_{CO_2}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuCO_3}=\dfrac{0,15.124}{20}.100\%=93\%\\\%m_{CuCl_2}=7\%\end{matrix}\right.\)
c, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
\(5.a.V_{rượu}=\dfrac{46.25}{100}=11,5\left(l\right)\\ m_{rượu}=11,5.0,8=9,2\left(g\right)\\ b.C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\\n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\\ n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,2\left(mol\right)\\ \Rightarrow m_{CH_3COOC_2H_5}=0,2.88=17,6\left(g\right)\\ VìH=30\%\Rightarrow m_{CH_3COOC_2H_5}=17,6.30\%=5,28\left(g\right)\)
\(6.a.C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ n_{CH_3COOH}=0,5.60=30\left(g\right)\\ b.n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,5\left(mol\right)\\ \Rightarrow m_{CH_3COOC_2H_5}=0,5.88=44\left(g\right)\\ VìH=70\%\Rightarrow m_{CH_3COOC_2H_5}=44.70\%=30,8\left(g\right)\)
Caai 7 :
a) C2H4 + Br2 $\to$ C2H4Br2
b) Theo PTHH : n C2H4 = n Br2 = 8/160 = 0,05(mol)
%V C2H4 = 0,05.22,4/2,24 .100% = 50%
%V CH4 = 100% -50% = 50%
Câu 8 :
a) C2H5OH = a(mol) => n CH3COOH = 2a(mol)
$C_2H_5OH + Na \to C_2H_5OH + \dfrac{1}{2}H_2$
$CH_3COOH + Na \to CH_3COONa + \dfrac{1}{2}H_2$
Theo PTHH :
n H2 = 1/2 n C2H5OH + 1/2 n CH3COOH = 0,5a + a = 3,36/22,4 = 0,15
=> a = 0,1
=> m = 0,1.46 + 0,1.2.60 = 16,6(gam)
b)
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
Ta thấy : n C2H5OH < n CH3COOH nên hiệu suất tính theo số mol C2H5OH
n CH3COOC2H5 = n C2H5OH pư = 0,1.80% = 0,08(mol)
m este = 0,08.88 = 7,04(gam)