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PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
a+b) Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{FeCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,6\cdot36,5}{14,6\%}=150\left(g\right)\\m_{FeCl_3}=0,2\cdot162,5=32,5\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{FeCl_3}=\dfrac{32,5}{150+16}\cdot100\%\approx19,58\%\)
b) PTHH: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PTHH: \(n_{KOH}=n_{HCl}=0,6\left(mol\right)\) \(\Rightarrow V_{KOH}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)
Bài 2. Cho 8g Fe2O3 tác dụng vừa đủ với dd HCl 20% (D = 1,1g/ml). Hãy tính: a. Thể tích dd HCl đã dùng b. Nồng độ % dd thu được sau phản ứng
a) \(n_{Fe_2O_3}=0,05\left(mol\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(n_{HCl}=6n_{Fe_2O_3}=0,3\left(mol\right)\)
=> \(m_{ddHCl}=\dfrac{0,3.36,5}{20\%}=54,75\left(g\right)\)
=> \(V_{HCl}=\dfrac{m}{D}=\dfrac{54,75}{1,1}=49,77\left(g\right)\)
b) \(m_{ddsaupu}=8+54,75=62,75\left(g\right)\)
\(C\%_{FeCl_3}=\dfrac{0,05.2.162,5}{62,75}.100=25,9\%\)
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)
a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)
b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)
c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
Ta có: \(n_{HCl}=0,2\cdot3=0,6\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{FeCl_3}=0,2\left(mol\right)\\n_{Fe_2O_3}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,1\cdot160=16\left(g\right)\\m_{FeCl_3}=0,2\cdot162,5=32,5\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_3}=\dfrac{32,5}{16+200\cdot1,1}\cdot100\%\approx13,77\%\\C_{M_{FeCl_3}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
\(a,PTHH:Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\\ b,n_{Ca\left(OH\right)_2}=\dfrac{7,4}{74}=0,1\left(mol\right)\\ \Rightarrow n_{HCl}=2n_{Ca\left(OH\right)_2}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{200}\cdot100\%=3,65\%\\ c,CaCl_2+H_2SO_4\rightarrow CaSO_4+2HCl\\ n_{H_2SO_4}=1\cdot0,25=0,25\left(mol\right)\\ n_{CaCl_2}=n_{Ca\left(OH\right)_2}=0,1\left(mol\right)\)
Do đó sau p/ứ H2SO4 dư
\(\Rightarrow n_{CaSO_4}=n_{CaCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{CaSO_4}=0,1\cdot136=13,6\left(g\right)\)
Ca(OH)2+ H2CL-> CaCL2+ H2O
số n của Ca(OH)2 là :
A) nCa(OH)2 =m/M=7,4/74=0,1 mol
ta có nCa(OH)2=nCaCL2=0,1 mol
=>mCaCL2=0,1.111=11,1 gam
B) số mol của HCL là
nHCL=nCa(OH).2=0,1.2=0,2 mol
khối lượng của dung dịch HCL cần dùng
mHCL=n.M=0,2.71=14,2 gam
C)
nồng độ phần trăm là :
C/.=11,1/214,6.100/.=5/.
\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,4 0,4 0,4 0,4
a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)
b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)
\(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)
\(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)
\(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)
c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)
0,4 0,3 0,3 0,3
\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)
\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)
Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)
2 3 1 3
0,8 1,2 0,4 1,2
a) \(n_{H2}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=1,2.22,4=26,88\left(l\right)\)
b) \(n_{H2SO4}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)
⇒ \(m_{H2SO4}=1,2.98=117,6\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{117,6.100}{29,4}=400\left(g\right)\)
c) \(n_{Al2\left(SO4\right)3}=\dfrac{1,2.1}{3}=0,4\left(mol\right)\)
⇒ \(m_{Al2\left(SO4\right)3}=0,4.342=136,8\left(g\right)\)
\(m_{ddspu}=21,6+400-\left(1,2.2\right)=419,2\left(g\right)\)
\(C_{Al2\left(SO4\right)3}=\dfrac{136,8.100}{419,2}=32,63\)0/0
Chúc bạn học tốt
BaO+H2O -> Ba(OH)2
0,02 0,02
a) CM = n/V = 0,02/0,02 = 1M
b) Ba(OH)2 + H2SO4 -> BaSO4 +2H2O
0,02 0,02
=> m = 0,392 g
D = m/V = 1,14
=> 0,392/V = 1,14 => V = 0,34l