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PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{Fe}=\dfrac{3,36}{56}=0,06\left(mol\right)\)
a, Theo PT: \(n_{FeCl_2}=n_{Fe}=0,06\left(mol\right)\Rightarrow m_{FeCl_2}=0,06.127=7,62\left(g\right)\)
b, Theo PT: \(n_{H_2}=n_{Fe}=0,06\left(mol\right)\Rightarrow V_{H_2}=0,06.22,4=1,344\left(l\right)\)
c, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT: \(n_{Cu\left(LT\right)}=n_{H_2}=0,06\left(mol\right)\)
\(\Rightarrow m_{Cu\left(LT\right)}=0,06.64=3,84\left(g\right)\)
Mà: mCu (TT) = 2,88 (g)
\(\Rightarrow H\%=\dfrac{2,88}{3,84}.100\%=75\%\)
PT: ��+2���→����2+�2Fe+2HCl→FeCl2+H2
Ta có: ���=3,3656=0,06(���)nFe=563,36=0,06(mol)
a, Theo PT: �����2=���=0,06(���)⇒�����2=0,06.127=7,62(�)nFeCl2=nFe=0,06(mol)⇒mFeCl2=0,06.127=7,62(g)
b, Theo PT: ��2=���=0,06(���)⇒��2=0,06.22,4=1,344(�)nH2=nFe=0,06(mol)⇒VH2=0,06.22,4=1,344(l)
c, PT: ���+�2��→��+�2�CuO+H2toCu+H2O
Theo PT: ���(��)=��2=0,06(���)nCu(LT)=nH2=0,06(mol)
⇒���(��)=0,06.64=3,84(�)⇒mCu(LT)=0,06.64=3,84(g)
Mà: mCu (TT) = 2,88 (g)
⇒�%=2,883,84.100%=75%⇒H%=3,842,88.100%=75%
2) PTHH: Zn +2HCl \(\rightarrow\) ZnCl2 + \(H_2\uparrow\)
a) nZn = \(\frac{19,5}{65}=0,3\left(mol\right)\)
Theo PT: n\(ZnCl_2\) = nZn = 0,3 (mol)
=> m\(ZnCl_2\) = 0,3.136 = 40,8 (g)
b) Theo PT: nHCl = 2nZn =2.0,3 = 0,6 (mol)
=> mHCl = 0,6.36,5 = 21,9 (g)
=> mdd HCl = \(\frac{21,9}{20}.100\) = 109,5 (g)
1) PTHH: Fe + 2HCl \(\rightarrow FeCl_2+H_2\uparrow\)(1)
a) nFe = \(\frac{2,8}{56}=0,05\left(mol\right)\)
Theo PT(1): n\(FeCl_2\) = nFe = 0,05 (mol)
=> m\(FeCl_2\) = 0,05.127 = 6,35 (g)
b) Theo PT(1): nHCl = nFe = 0,05(mol)
=> mHCl = 0,05.36,5 = 1,825 (g)
=> mdd HCl = \(\frac{1,825.100}{20}=9,125\left(g\right)\)
c) PTHH: 2xM + 2yHCl \(\rightarrow\) 2MxCly + yH2\(\uparrow\)(2)
Theo PT(1): n\(H_2\) = nFe = 0,05 (mol) = n\(H_2\)(2)
Theo PT(2): nM =\(\frac{2x}{y}n_{H_2}\) = \(\frac{2x}{y}.0,05=\frac{0,1x}{y}\)(mol)
=> MM = \(\frac{1,2}{\frac{0,1x}{y}}=\frac{12y}{x}\)(g/mol)
Ta có bảng sau:
| x | 1 | 2 | 3 |
| y | 2 | 3 | 4 |
| M | 24 | 18 | 16 |
| Mg | loại | loại |
Vậy M là magie (Mg)
a) \(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
b) \(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH, ta có:
\(n_{HCl}=2n_{H_2}=2.0,15=0,3\left(mol\right)\)
\(m_{HCl}=n_{HCl}.M_{HCl}=0,3.36,5=10,95\left(g\right)\)
c) Theo PTHH, ta có:
\(n_{FeCl_2}=n_{H_2}=0,15\left(mol\right)\)
\(m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,15.127=19,05\left(g\right)\)
a) PTHH: Fe + 2HCl ➞ FeCl2 + H2 (1)
b) nFe= \(\dfrac{14}{56}=0,25\) (mol)
Theo PT: \(n_{FeCl_2}=n_{Fe}=0,25\) (mol)
⇒ \(m_{FeCl_2}\)= 0,25 . 127 = 31,75 (g)
c) PTHH khử của H2: H2 + CuO ➞ Cu + H2O
Theo PT (1): \(n_{H_2}=n_{Fe}=0,25\) (mol)
Theo PT (2): \(n_{Cu}=n_{H_2}=0,25\) (mol)
⇒ mCu= 0,25 . 64 = 16 (g)
( Phần c) không tính Cu theo CuO vì CuO dư )
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
A. \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
Theo PTHH: \(n_{H_2}=n_{Fe}=0,4\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)
B. Theo PTHH: \(n_{FeCl_2}=n_{Fe}=0,4\left(mol\right)\)
\(m_{FeCl_2}=0,4.127=50,8\left(g\right)\)
C. Nồng độ mol:
\(C_M=\dfrac{0,4}{0,3}=1,3\left(M\right)\)
Fe+2HCl\(\rightarrow\)FeCl2+H2
\(n_{Fe}=\dfrac{14}{56}=0,25mol\)
\(n_{H_2}=n_{Fe}=0,25mol\)
\(n_{FeCl_2}=n_{Fe}=0,25mol\)
\(m_{FeCl_2}=0,25.127=31,75gam\)
H2+CuO\(\overset{t^0}{\rightarrow}\)Cu+H2O
\(n_{CuO}=\dfrac{25,6}{80}=0,32mol\)
Tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,32}{1}\)\(\rightarrow\)CuO dư
\(n_{Cu}=n_{CuO}=n_{H_2}=0,25mol\)
\(m_{Cu}=0,25.64=16gam\)
\(n_{CuO\left(dư\right)}=0,32-0,25=0,07mol\)
\(m_{CuO}=0,07.80=5,6gam\)
a) Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b) nFe = \(\dfrac{11,2}{56}=0,2mol\)
Theo pt: nH2 = nFe = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48lit
c) Theo pt: nHCl = 2nFe = 0,4 mol
=> mHCl = 0,4.36,5 = 14,6 g
=> C% = \(\dfrac{14,6}{73}.100\%=20\%\)
a) Pt: Fe+2HCl→FeCl2+H2Fe+2HCl→FeCl2+H2
b) nFe = \(\dfrac{11,2}{56}=0,2mol\)
Theo pt: nH2 = nFe = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48lit
c) Theo pt: nHCl = 2nFe = 0,4 mol
=> mHCl = 0,4.36,5 = 14,6 g
=> \(C\%=\dfrac{14,6}{73}.100\%=20\%\)
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