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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
PTHH:
2Al + 6HCl ---> 2AlCl3 + 3H2 (1)
Al2O3 + 6HCl ---> 2AlCl3 + 3H2 (2)
Theo PT(1): \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,06=0,04\left(mol\right)\)
\(\Rightarrow m_{Al}=0,04.27=1,08\left(g\right)\)
\(\Rightarrow m_{Al_2O_3}=2,1-1,08=1,02\left(g\right)\)
\(\Rightarrow n_{Al_2O_3}=\dfrac{1,02}{102}=0,01\left(mol\right)\)
Theo PT(1): \(n_{HCl}=3.n_{Al}=3.0,04=0,12\left(mol\right)\)
Theo PT(2): \(n_{HCl}=6.n_{Al_2O_3}=6.0,01=0,06\left(mol\right)\)
\(\Rightarrow m_{HCl}=\left(0,06+0,12\right).36,5=6,57\left(g\right)\)
Ta có: \(C_{\%_{HCl}}=\dfrac{6,57}{m_{dd_{HCl}}}.100\%=7,3\%\)
\(\Rightarrow m_{dd_{HCl}}=90\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{NaCl}=0,2.0,2=0,04\left(mol\right)\\ m_{NaCl}=0,04.58,5=2,34\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2\)
\(n_{Al}= \dfrac{5,4}{27}= 0,2 mol\)
Theo PTHH:
\(n_{AlCl_3}= n_{Al}= 0,2 mol\)
\(\Rightarrow m_{AlCl_3}= 0,2 . 133,5=26,7 g\)
Theo PTHH:
\(n_{H_2}= \dfrac{3}{2} n_{Al}= 0,3 mol\)
\(\Rightarrow V= 0,3 . 22,4= 6,72 l\)
b)
Theo PTHH:
\(n_{HCl}= 3n_{Al}= 0,6 mol\)
\(\Rightarrow m_{HCl}= 0,6 . 36,5=21,9 g\)
\(\Rightarrow m_{dd HCl}= \dfrac{21,9 . 100}{15}= 146 g\) ( nếu ở tử là : 21,9 . 100% thì ở mẫu bạn chia cho 15% nhé)
![](https://rs.olm.vn/images/avt/0.png?1311)
2al+ 6hcl-> 2alcl3+3h2
a-> 3a a 1,5a
fe+2hcl-> fecl2+h2
b->2b b b
27a+56b= 5,5
1,5a+b=4,48/22,4
=> a=0,1; b=0,05
=> %mal=0,1*27/5,5*100=49,09%
=>%mfe= 100-49,09=50,9%
mhcl= 3a+2b= 3*0,1+2*0,05=0,4
=>mddhcl= 0,4*36,5*100/14,6=100g
-> vddhcl=100/ 1,08=92,592ml
mddsau pư= 5,5+100-0,2*2=105,1
C% alcl3= 133,5*0,1/105,1*100=12,7
Cfecl2= 127* 0,05/105,1*100=6,04
\(n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ 2CH_3COOH + Zn \to (CH_3COO)_2Zn + H_2\\ n_{CH_3COOH} = 2n_{Zn} = 0,2(mol)\\ m_{CH_3COOH} = 0,2.60 = 12(gam)\)
Đáp án B.