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a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,1 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{AlCl_3}=0,1.133,5=13,35g\)
d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,15 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
a.\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,2 0,3 ( mol )
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b.\(n_{CuO}=\dfrac{48}{80}=0,6\left(mol\right)\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,6 > 0,3 ( mol )
0,3 0,3 ( mol )
\(m_{Cu}=0,3.64=19,2\left(g\right)\)
c.\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,3 0,15 ( mol )
\(V_{kk}=V_{O_2}.5=\left(0,15.22,4\right).5=16,8\left(l\right)\)
a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(2mol\) \(6mol\) \(2mol\) \(3mol\)
\(0,27\) \(x\) \(y\) \(z\)
b) ta có: \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{7,3}{27}=0,27\left(mol\right)\)
theo PT: \(n_{Al}=n_{AlCl_3}=0,27\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,27.133,5=36,045\left(g\right)\)
c) ta có: \(n_{H_2}=\dfrac{m_{H_2}}{M_{H_2}}=\) \(\dfrac{0,27.3}{2}=0,405\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=n_{H_2}.22,4=0,405.22,4=9,072\left(l\right)\)
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) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2Al + 6HCl ---> 2AlCl3 + 3H2
Theo PTHH: \(n_{AlCl_3}=n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\)
=> mAl = 0,4.27 = 10,8 (g)
b) \(m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
a.\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,2 0,3 ( mol )
\(m_{AlCl_3}=0,2.133,5=26,7g\)
\(V_{H_2}=0,3.22,4=6,72l\)
b.\(n_{CuO}=\dfrac{56}{80}=0,7mol\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,7 < 0,3 ( mol )
0,3 0,3 0,3 ( mol )
\(m_X=m_{CuO\left(dư\right)}+m_{Cu}=\left[\left(0,7-0,3\right).80\right]+\left(0,3.64\right)=51,2g\)
nAl=\(\dfrac{5,8}{27}\)≈0,215 mol
a, PTPƯ: 2Al + 6HCl ---> 2AlCl3 + 3H2
Ta có: 2 mol Al ---> 3 mol H2
nên 0,215 mol Al ---> 0,323 mol H2
=> VH2=0,323.22,4≈7,24 l
b, Ta có: 2 mol Al ---> 6 mol HCl
nên 0,215 mol Al ---> 0,65 mol HCl
=> VHCl=0,65.22,4=14,56 l
c, Ta có: 2 mol Al ---> 2 mol AlCl3
nên 0,215 mol Al ---> 0,215 mol AlCl3
=> mAlCl3=0,215.133,5≈28,7 g
nAl = 2,7 : 27 = 0,1 (mol)
pthh : 2Al + 6HCl -> 2AlCl3 + 3H2
0,1 0,3 0,1 0,15
=> mAlCl3 = 0,1 . 133,5 = 13,35 (G)
=> VH2 = 0,15 . 22,4 = 3,36 (L)
pthh : CuO + H2 -t-> H2O + Cu
0,15 0,15
=> mCuO = 0,15 . 64 = 9,6 (G)
a)nAl = 2,7/7=0,1(mol)
có pthh : 2Al + 6HCl -> 2AlCl3 + 3H2
0,1 0,3 0,1 0,15
Theo PT ta có: nHCl = 3nAl = 0,1 : 3 = 0,3(mol)
mHCl = 0,1 x 133,5 = 13,35(g)
b)=> VH2 = 0,15 . 22,4 = 3,36 (L)
có pthh : CuO + H2 -t-> H2O + Cu
0,15 0,15
=> mCuO = 0,15 . 64 = 9,6 (G)