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nH2 = 17.92/22.4 = 0.8 (mol)
nHCl = 2nH2 = 1.6 (mol)
BTKL :
mhh + mHCl = mM + mH2
=> 44.5 + 1.6*36.5 = mM + 0.8*2
=> mM = 101.3 (g)
2 Al + 6 HCl -> 2 AlCl3 + 3 H2
x_____3x______x_____1,5x(mol)
Mg + 2 HCl -> MgCl2 + H2
y__2y________y___y(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}27x+24y=44,5\\22,4.1,5x+22,4y=17,92\end{matrix}\right.\)
Anh nhìn đề sai chắc luôn
a)
Gọi số mol Mg, Al là a, b (mol)
=> 24a + 27b = 26,25 (1)
\(n_{H_2}=\dfrac{30,8}{22,4}=1,375\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
a-->2a--------->a------>a
2Al + 6HCl --> 2AlCl3 + 3H2
b---->3b------->b------>1,5b
=> a + 1,5b = 1,375 (2)
(1)(2) => a = 0,25 (mol); b = 0,75 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{26,25}.100\%=22,857\%\\\%m_{Al}=\dfrac{0,75.27}{26,25}.100\%=77,143\%\end{matrix}\right.\)
b)
nHCl = 2a + 3b = 2,75 (mol)
=> mHCl = 2,75.36,5 = 100,375 (g)
=> \(m_{dd.HCl}=\dfrac{100,375.100}{10}=1003,75\left(g\right)\)
c)
mdd sau pư = 1003,75 + 26,25 - 1,375.2 = 1027,25 (g)
\(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,25.95}{1027,25}.100\%=2,312\%\\C\%_{AlCl_3}=\dfrac{0,75.133,5}{1027,25}.100\%=9,747\%\end{matrix}\right.\)
2al+6hcl-> 2alcl3+ 3h2
fe+2hcl-> fecl2+h2
nh2=13,44/22,4=0,6 mol
27a+56b=16,5
1,5a+ b=0,6
a=0,3, b=0,15
%mal=0,3*27/16,5*100=49,09%
%mfe=50,9%
nhcl=3a+2b=1,2
Vdd hcl=1,2/2=0,6l
\(n_{H_2}=\frac{V_{H_2}}{224,}=\frac{1,68}{22,4}=0,075\left(mol\right)\)
-> \(m_{H_2}=n_{H_2}.M_{H_2}=0,075.2=0,15\left(g\right)\)
Mà \(n_{\left(H\right)}=2n_{H_2}\)
-> \(n_{\left(H\right)}=2n_{H_2}=2.0,075=0,15\left(mol\right)\)
-> \(n_{HCl}=n_{\left(H\right)}=0,15\left(mol\right)\)
-> \(m_{HCl}=n_{HCl}.M_{HCl}=0,15.\left(1+35,5\right)=5,475\left(g\right)\)
- Áp dụng định luật bảo toàn khối lượng .
\(m_{hh}+m_{HCl}=m_{H_2}\) + mMuối khan
=> mMuối khan = \(m_{hh}+m_{HCl}-m_{H_2}=2,17+5,475-0,15\)
=> mMuối khan = 7,495 ( g )
\(n_{H_2}=\frac{1,68}{22,4}=0,075\left(mol\right)\)
\(m_{H_2}=0,075.2=0,15\left(g\right)\)
\(n_{HCl}=2.n_{H_2}=2.0,075=0,15\left(mol\right)\)
\(m_{HCl}=0,15.36,5=5,475\left(g\right)\)
Theo DDLBTKL thì:
\(m_{muối}=2,17+5,475-0,15=7,495\left(g\right)\)
Chúc bạn học tốt