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a.b.
\(n_{Fe}=\dfrac{2,8}{56}=0,05mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,1 0,05 ( mol )
\(V_{H_2}=0,05.22,4=1,12l\)
\(m_{HCl}=0,1.36,5=3,65g\)
c.
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,05 0,05 ( mol )
\(m_{Cu}=0,05.64=3,2g\)
`a)`
PTHH : `Fe + 2HCl -> FeCl_2 + H_2`
`b)`
`n_{Fe} = (11,2)/(56) = 0,2` `mol`
`n_{HCl} = 2 . n_{Fe} = 0,4` `mol`
`m_{HCl} = 0,4 . 36,5 = 14,6` `gam`
`c)`
`n_{FeCl_2} = n_{Fe} = 0,2` `mol`
`m_{FeCl_2} = 0,2 . 127 = 25,4` `gam`
`n_{H_2} = n_{Fe} = 0,2` `mol`
`V_{H_2} = 0,2 . 22,4 = 4,48` `l`
a) $Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH :
$n_{H_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$V_{H_2} = 0,2.22,4 = 4,48(lít)$
b) $n_{HCl} = 2n_{Fe} = 0,4(mol)$
$m_{HCl} = 0,4.36,5 = 14,6(gam)$
c) $2H_2 + O_2 \xrightarrow{t^o}2H_2O$
Theo PTHH :
$V_{O_2} = \dfrac{1}{2}V_{H_2} = 2,24(lít)$
$n_{H_2O} = n_{H_2} = 0,2(mol)$
$m_{H_2O} = 0,2.18 = 9(gam)$
\(a,\text{Sơ đồ p/ứ: }Fe+HCl\to FeCl_2+H_2\\ b,PTHH:Fe+2HCl\to FeCl_2+H_2\\ c,\text{Bảo toàn KL: }m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}+56=150+8=158\\ \Rightarrow m_{HCl}=102(g)\)
BTKL: \(m_{Fe}+m_{HCl}=m_{muối}+m_{H_2}\)
\(\Rightarrow m_{H_2}=5,6+7,3-12,7=0,2\left(g\right)\)
a) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5-------1---------0,5------0,5
b) \(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)
c) \(H_2+CuO\rightarrow Cu+H_2O\)
0,5-----0,5------0,5----0,5
Khối lượng đồng tạo thành: \(m_{Cu}=n_{Cu}.64=0,5.64=32\left(g\right)\)
a) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
0,5-------------------------->0,5`
b) `V_{H_2} = 0,5.22,4 = 11,2 (l)`
c) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,5---->0,5
`=> m_{Cu} = 0,5.64 = 32 (g)`
\(Fe+2HCl\underrightarrow{t^o}FeCl_2+H_2\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(V_{H_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)
\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(m_{Cu}=n.M=0,5.64=32\left(g\right)\)
a, \(PTHH:CuO+H_2\underrightarrow{^{t^o}}Cu+H_2O\)
Ta có:
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow n_{Cu}=n_{H2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)
b,
i .\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
ii. \(n_{FeCl2}=116,2\left(g\right)\) ( Đề cho)
iii: \(n_{FeCl2}=\frac{116,2}{127}=0,9\left(mol\right)\)
\(\Rightarrow n_{HCl}=2n_{FeCl2}=1,8\left(mol\right)\)
\(\Rightarrow m_{HCl}=1,8.36,5=65,7\left(g\right)\)
Fe+2HCl\(\rightarrow\)FeCl2+H2
-Áp dụng định luật bảo toàn khối lượng:
\(m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\)
\(\rightarrow\)\(m_{HCl}=m_{FeCl_2}+m_{H_2}-m_{Fe}=254+4-112=146gam\)