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a. 2Al + 6HCl -> 2AlCl3 + 3H2
b. nAl = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)
\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)
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) \(m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
\(n_{Al}=\dfrac{21.06}{27}=0.78\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.78..............................1.17\)
\(V_{H_2}=1.17\cdot22.4=26.208\left(l\right)\)
2Al+6HCl->2AlCl3+3H2
0,4--------------------------0,6
n Al=0,4 mol
=>VH2=0,6.22,4=13,44l
H2+HgO-tO>Hg+H2O
0,6--------------0,6
=>m Hg=0,6.201=120,6g
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,4 0,6
\(\rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\\ PTHH:HgO+H_2\underrightarrow{t^o}Hg+H_2P\)
0,6 0,6
\(\rightarrow m_{Hg}=0,6.201=120,6\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
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
\(m_{Al\left(dư\right)}=\left(0,2-\dfrac{0,4}{3}\right).27=1,8\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,2 0,2 0,3
\(V_{H_2}=n.22,4=6,72\left(l\right)\)
\(m_{AlCl_3}=n.M=0,2.133,5=26,7\left(g\right)\)
18,25 là số gam của dd mà sao tính đc công thức đấy , dd tính theo công thức n/V thôi chứ .
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)
d, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Có: \(\dfrac{0,2}{1}>\dfrac{0,45}{3}\) → Fe2O3 dư.
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
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: \(n_{H_2}=\dfrac{m_{H_2}}{M_{H_2}}=\) \(\dfrac{0,27.3}{2}=0,405\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=n_{H_2}.22,4=0,405.22,4=9,072\left(l\right)\)
7,3 là KL của axit mà em