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\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
a: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 0,2
b: V=0,2*22,4=4,48(lít)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
a) Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{Fe}\)
\(\Rightarrow m_{Fe}=0,2\cdot56=11,2\left(g\right)\) \(\Rightarrow m_{Fe_2O_3}=16\left(g\right)\)
b+c) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=0,2\left(mol\right)\\n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=2n_{Fe}+6n_{Fe_2O_3}=1\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{36,5}{20\%}=182,5\left(g\right)\)
Mặt khác: \(n_{FeCl_2}=0,2\left(mol\right)=n_{H_2}=n_{FeCl_3}\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\m_{FeCl_3}=0,2\cdot162,5=32,5\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{hhA}+m_{ddHCl}-m_{H_2}=209,3\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{25,4}{209,3}\cdot100\%\approx12,14\%\\C\%_{FeCl_3}=\dfrac{32,5}{209,3}\cdot100\%\approx15,53\%\end{matrix}\right.\)
\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)
1 6 2 3
0,1 0,6 0,2
a) \(n_{Fe}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(m_{Fe2O3}=27,2-11,2=16\left(g\right)\)
b) Có : \(m_{Fe2O3}=16\left(g\right)\)
\(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,4+0,6=1\left(mol\right)\)
⇒ \(m_{HCl}=1.36,5=36,5\left(g\right)\)
\(m_{ddHCl}=\dfrac{36,5.100}{20}=182,5\left(g\right)\)
c) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{FeCl2}=0,2.127=25,4\left(g\right)\)
\(n_{FeCl3}=\dfrac{0,6.2}{6}=0,2\left(mol\right)\)
⇒ \(m_{FeCl3}=0,2.162,5=32,5\left(g\right)\)
\(m_{ddspu}=27,2+182,5-\left(0,2.2\right)=209,3\left(g\right)\)
\(C_{FeCl2}=\dfrac{25,4.100}{209,3}=12,14\)0/0
\(C_{FeCl3}=\dfrac{32,5.100}{209,3}=15,53\)0/0
Chúc bạn học tốt
\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(V_{H_2}=\left(0,1+0,2\right).22,4=6,72l\\ b)V_{ddHCl}=\dfrac{0,2+0,4}{2}=0,3l\\ c)m_{muối}=0,1.127+95.0,2=31,7g\)
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{FeSO_4}=n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,2.152=30,4\left(g\right)\)
c, \(C_{M_{H_2SO_4}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
_____0,1-->0,2----->0,1----->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
b) \(V_{ddHCl}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
c) \(C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,5}=0,2M\)
@@ bạn nên đăng từng câu một thôi , nhìn vào ngại giải lắm
===============
Bài 11 :
a, PTHH :
CaO + 2HCl -> CaCl2 + H2O
x...............2x..............x (mol)
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
y..................2y...........y........................y (mol)
Gọi x,y lần lượt là số mol CaO , CaCO3 trong hỗn hợp ( x,y>0)
Khi đó nCO2=y= \(\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
mmuoi khan = mCaCl2 = 111(x+y) =66,6(g)
=> x+y=0,6 (mol) => x=0,15(mol)
\(\Rightarrow\left\{{}\begin{matrix}m_{CaO}=0,15.56=8,4\left(g\right)\\m_{CaCO_3}=0,6.100=60\left(g\right)\end{matrix}\right.\)
=> mX = 68,4 (g)
=> \(\left\{{}\begin{matrix}\%m_{CaO}=\dfrac{8,4}{68,4}.100\%=12,28\%\\\%m_{CaCO_3}=87,72\%\end{matrix}\right.\)
b , \(\Sigma_{HCl}=2\left(x+y\right)=2.0,6=1,2\left(mol\right)\)
=> mHCl = 43,8(g) => mdung dịch HCl = 43,8:14,6%=300(g)
=>V=\(\dfrac{m}{D}=\dfrac{300}{1,12}\approx267,86\left(ml\right)\)
c, mB = mX + mdung dich HCl - mCO2 = 68,4+300-0,45.44 =348,6(g)
=> C%B = \(\dfrac{66,6}{348,6}.100\%\approx19,105\%\)
Câu 13 dd thu đc là j nx v
E nghĩ NaOH+HCl-> NaCl+H2O
Còn j nx??? NaCl có pứ với Al2O3 đâu
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)
c, \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_2}=0,2.127=25,4\left(g\right)\)