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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) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
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) \(m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
2Al+6HCl->2AlCl3+3H2
0,05----0,15------------------0,075 mol
n H2=\(\dfrac{1,68}{22,4}\)=0,075 mol
=>m Al=0,05.27=1,35g
=>HCl dư =>m HCl=0,1.36,5=3,65g
\(a) 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ b) n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)\\ m_{Al} = 0,2.27 = 5,4(gam)\\ c) n_{HCl\ pư} = 2n_{H_2} = 0,6(mol)\\ n_{HCl\ đã\ dùng} = \dfrac{0,6}{80\%} = 0,75(mol)\\ m_{dd\ HCl} = \dfrac{0,75.36,5}{54,75\%} = 50(gam)\)
a)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,2----------->0,2----->0,3
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b) \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
c)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
0,3<----------------0,3
=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
\(a,n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2--------------->0,2------->0,3
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ b,m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
c, PTHH:
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,2<------------------0,2
\(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)
a, PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\)
b, Ta có: \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=6n_{Al_2O_3}=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,2 0,3 ( mol )
\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)
\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,2.133,5=26,7g\)
a) PTHH: 2Al + 6HCl -> 2AlCl3 + 3 H2
b) nHCl=0,6(mol); nAl=0,3(mol)
Ta có: 0,3/2 > 0,6/6
=> HCl hết, Al dư, tính theo nHCl
c) nH2= 3/6 . nHCl=3/6 . 0,6= 0,3(mol)
=> V=V(H2,đktc)=0,3.22,4= 6,72(l)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
a) 2Al + 6HCl --> 2AlCl3 + 3H2
b) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
2Al + 6HCl --> 2AlCl3 + 3H2
\(\dfrac{0,4}{3}\)<--------------------0,2
=> Al dư
\(m_{Al\left(dư\right)}=\left(0,2-\dfrac{0,4}{3}\right).27=1,8\left(g\right)\)