Hòa tan Fe phải dùng hết 200g dung dịch HCl 7.3%.
a) tính mFe và C% FeCl2
b) Lượng khí H2 thoát ra khử dược ba nhiêu g CuO
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PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a) Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)=n_{H_2}\) \(\Rightarrow V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
b) Theo PTHH: \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{0,4\cdot36,5}{200}\cdot100\%=7,3\%\)
b) Theo PTHH: \(n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{ddHCl}-m_{H_2}=210,8\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{210,8}\cdot100\%\approx12,05\%\)
a) n Fe = 28/56 = 0,5(mol)
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH :
n HCl = 2n Fe = 1(mol)
=> m dd HCl = 1.36,5/10% = 365(gam)
b)
n FeCl2 = n H2 = n Fe = 0,5(mol)
Suy ra :
V H2 = 0,5.22,4 = 11,2(lít)
m FeCl2 = 0,5.127 = 63,5(gam)
c)
Sau phản ứng:
mdd = m Fe + mdd HCl - m H2 = 28 + 365 - 0,5.2 = 392(gam)
=> C% FeCl2 = 63,5/392 .100% = 16,2%
\(a.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ b.n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=127.0,1=12,7\left(g\right)\\ c.V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
a ) PTHH : Zn +2HCl -> ZnCl2 +H2
TA CÓ :nZn=6,5 /65=0,1 mol
Zn+2HCl ->ZnCl2 +H2
THEO ĐB:0,1 a
THEO PT:1 1
nH2=a=0,1 .1 :1=0,1 mol
=>THỂ TÍCH H2=0,1.22,4 =2,24 (l)
\(a)Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{HCl}=\dfrac{200.18,25\%}{100\%.36,5}=1mol\\ n_{FeCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=1:2=0,5mol\\ m_{FeCl_2}=0,5.127=63,5g\\ c)V_{H_2}=0,5.24,79=12,395l\)
\(C\%_{ddHCl}=\dfrac{m_{HCl}}{m_{ddHCl}}.100\%\)
\(\Leftrightarrow m_{HCl}=\dfrac{C\%_{ddHCl}.m_{ddHCl}}{100\%}\)
\(\Leftrightarrow m_{HCl}=\dfrac{18,25\%.200}{100\%}\)
\(\Rightarrow m_{ddHCl}=36,5g\)
\(n_{HCl}=\dfrac{m}{M}=\dfrac{36,5}{36,5}=1mol\)
PTHH: Fe + 2HCl \(\rightarrow\) FeCl2 + H2
TL: 1 2 1 1
mol: 0,5 \(\leftarrow\) 1 \(\rightarrow\) 0,5 \(\rightarrow\) 0,5
\(a.m_{FeCl_2}=n.M=0,5.127=63,5g\)
\(c.V_{H_2}=n.22,4=0,5.22,4=11,2l\)
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
PTHH :
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,1 0,05 0,05
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
\(V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,1}{1}=0,1\left(l\right)\)
\(m_{FeCl_2}=0,05.127=6,35\left(g\right)\)
\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\
n_{HCl}=0,5.1=0,5mol\\
Fe+2HCl\rightarrow FeCl_2+H_2\\
\Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\
Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95g\\ b)m_{FeCl_2}=0,1.127=12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl\left(dư\right)}=\dfrac{10,95}{200}\cdot100=5,475\%\\ C_{\%HCl\left(pư\right)}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,3 0,6 0,3 ( mol )
\(m_{Fe\left(pứ\right)}=0,3.56=16,8g\)
\(m_{Fe\left(dư\right)}=28-16,8=11,2g\)
\(m_{HCl}=0,6.36,5=21,9g\)
Fe + 2HCl \(\rightarrow\) FeCl2 + H2 (1)
a) mHCl = \(\dfrac{m_{dd}.C\%}{100\%}=\dfrac{200.7,3\%}{100\%}=14,6\left(g\right)\)
=> nHCl = 14,6 : 36,5 = 0,4(mol)
Theo PT (1)=> nFe =1/2 .nHCl = 1/2 . 0,4 = 0,2(mol)
=> mFe = 0,2 . 56 = 11,2(g)
Theo PT(1) => nH2 = nFe = 0,2(mol)
=> mH2 = 0,2 .2 = 0,4(g)
Theo ĐLBTKL :
mdd sau pứ = mFe + mdd HCl - mH2 = 11,2 + 200 - 0,4 =210,8(g)
Theo PT(1) => nFeCl2 = nFe = 0,2(mol)
=> mFeCl2 = 0,2 .127 =25,4(g)
=> C%FeCl2 = \(\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{25,4}{210,8}.100\%=12,05\%\)
b) H2 + CuO \(\underrightarrow{t^o}\) Cu +H2O (2)
Theo PT(2) => nCuO = nH2 = 0,2(mol)
=> mCuO = 0,2 . 80 =16(g)
khối lượng chất tan Fe cần dùng là
mct=\(\dfrac{C\%.mdd}{100\%}\)
mctFe=\(\dfrac{7,3.200}{100}\)=14,6g