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\(a.Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ n_{NaCl}=n_{HCl}=2.n_{CO_2}=2.\dfrac{448:1000}{22,4}=0,04\left(mol\right)\\ C_{MddHCl}=\dfrac{0,04}{0,02}=2\left(M\right)\\ b.m_{NaCl}=58,5.0,04=2,34\left(g\right)\\ c.m_{Na_2CO_3}=106.0,02=2,12\left(g\right)\\ \%m_{Na_2CO_3}=\dfrac{2,12}{5}.100=42,4\%\\ \%m_{NaCl}=100\%-42,4\%=57,6\%\)
Bài 16 :
\(n_{CO2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Pt : \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O|\)
1 2 2 1 1
0,02 0,04 0,04 0,02
a) \(n_{HCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)
20ml = 0,02l
\(C_{M_{HCl}}=\dfrac{0,04}{0,02}=2\left(M\right)\)
b) \(n_{NaCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)
⇒ \(m_{NaCl}=0,04.58,5=2,34\left(g\right)\)
c) \(n_{Na2CO3}=\dfrac{0,04.1}{2}=0,02\left(mol\right)\)
⇒ \(m_{Na2CO3}=0,02.106=2,12\left(g\right)\)
\(m_{NaCl}=5-2,12=2,88\left(g\right)\)
0/0Na2CO3 = \(\dfrac{2,12.100}{5}=42,4\)0/0
0/0NaCl = \(\dfrac{2,88.100}{5}=57,6\)0/0
Chúc bạn học tốt
\(n_{CuO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)
\(m=80a+160b=20\left(g\right)\left(1\right)\)
\(n_{HCl}=0.2\cdot3.5=0.7\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(n_{HCl}=2a+6b=0.7\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.05,b=0.1\)
\(m_{CuO}=0.05\cdot80=4\left(g\right)\)
\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
a, Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=1,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)
b, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
2Al + 6HCl \(\rightarrow\)2AlCl3 + 3H2 (1)
ZnO + 2HCl \(\rightarrow\)ZnCl2 + H2O (2)
nH2=\(\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Theo PTHH 1 ta có:
\(\dfrac{2}{3}\)nH2=nAl=0,4(mol)
mAl=0,4.27=10,8(g)
%mAl=\(\dfrac{10,8}{27}.100\%=40\%\)
%mZnO=100-40=60%
b;mZnO=27-10,8=16,2(g)
nZnO=\(\dfrac{16,2}{81}=0,2\left(mol\right)\)
Theo PTHH 1 và 2 ta có:
3nAl=nHCl(1)=1,2(mol)
2nZnO=nHCl(2)=0,4(mol)
mHCl=36,5.(0,4+1,2)=58,4(g)
mdd HCl=\(58,4:\dfrac{29,2}{100}=200\left(g\right)\)
c;
Theo PTHH 1 và 2 ta có:
nAl=nAlCl3=0,4(mol)
mAlCl3=0,4.133,5=53,4(g)
nZn=nZnCl2=0,2(mol)
mZnCl2=0,2.136=27,2(g)
C% dd AlCl3=\(\dfrac{53,4}{27+200-0,6.2}.100\%=23,65\%\)
C% dd ZnCl2=\(\dfrac{27,2}{27+200-1,2}.100\%=12,04\%\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
de: 0,4 \(\leftarrow\) 1,2 \(\leftarrow\) 0,4 \(\leftarrow\) 0,6
\(m_{Al}=0,4.27=10,8g\)
\(m_{ZnO}=27-10,8=16,2g\)
\(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)
a, \(\%m_{Al}=\dfrac{10,8}{27}.100\%=40\%\)
\(\%m_{ZnO}=100-40=60\%\)
ZnO + 2HCl \(\rightarrow\) ZnCl2 + H2O
de: 0,2 \(\rightarrow\) 0,4 \(\rightarrow\) 0,2
b, \(m_{HCl}=36,5.\left(1,2+0,2\right)=51,1g\)
\(m_{ddHCl}=\dfrac{51,1}{29,2}.100=175g\)
c, \(m_{ZnCl_2}=0,2.136=27,2g\)
\(m_{AlCl_3}=0,4.133,5=53,4g\)
\(m_{dd}=175+27-0,6.2=200,8g\)
\(C\%_{ZnCl_2}=\dfrac{27,2}{200,8}.100\%\approx13,55\%\)
\(C\%_{AlCl_3}=\dfrac{53,4}{200,8}.100\%\approx26,59\%\)
Đặt :
nCuO = x mol
nFe2O3 = y mol
mhh = 80x + 160y = 3.2 g (1)
CuO + 2HCl --> CuCl2 + H2O
x______2x_______x
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
y________6y______2y
x = 2y <=> x - 2y = 0 (2)
Giải (1) và (2) :
x = 0.02
y = 0.01
mCuO = 1.6 g
mFe2O3 = 1.6 g
mCuCl2 = 2.7 g
mFeCl3 = 3.25 g
nHCl = 2x + 6y = 0.1 mol
VHCl = 0.1/1 = 0.1 l
CM CuCl2 = 0.2 M
CM FeCl3 = 0.2 M