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\(a)\\ 4Al+ 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ \)
b) Bảo toàn khối lượng :
\(m_{O_2} = 21,8 -13,8 =8(gam)\\ n_{O_2} = \dfrac{8}{32} = 0,25(mol)\\ V_{O_2} = 0,25.22,4 = 5,6(lít)\)
c)
\(n_{Al} = a(mol) ; n_{Fe} = b(mol)\Rightarrow 27a + 56b = 13,8(1)\\ n_{O_2} = 0,75a + \dfrac{2}{3}b = 0,25(2)\\ (1)(2)\Rightarrow a = 0,2 ; b = 0,15\\ \%m_{Al} = \dfrac{0,2.27}{13,8}.100\% =39,13\%\\ \%m_{Fe} = 100\% -39,13\% = 60,87\%\)
a) Gọi số mol Al, Zn là 2a, a (mol)
PTHH: 4Al + 3O2 --to--> 2Al2O3
2a-->1,5a---------->a
2Zn + O2 --to--> 2ZnO
a---->0,5a------->a
=> \(102a+81a=18,3\)
=> a = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{0,2.27+0,1.65}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{0,2.27+0,1.65}.100\%=54,622\%\end{matrix}\right.\)
b) \(n_{O_2}=1,5a+0,5a=0,2\left(mol\right)\)
=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
\(m_{cr}=m_{Fe}=12.8\left(g\right)\)
\(NaOH+Al+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\)
\(m_{Al}=m_{hh}-m_{Fe}=18.2-12.8=5.4\left(g\right)\)
\(\%m_{Al}=\dfrac{5.4}{18.2}\cdot100\%=29.67\%\)
\(a) n_{Mg}= a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b =2,55(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{2,8}{22,4}=0,125(2)\\ (1)(2) \Rightarrow a = b = 0,05\\ \%m_{Mg} = \dfrac{0,05.24}{2,55}.100\% = 47,06\%\ ;\ \%m_{Al} =100\% -47,06\% = 52,94\%\\ b) n_{HCl} = 2n_{H_2} = 0,125.2 = 0,25(mol)\\ m_{dd\ HCl} = \dfrac{0,25.36,5}{7,3\%} = 125(gam)\\ V_{dd\ HCl} = \dfrac{125}{1,2} = 104,17(ml)\)
\(2CO + O_2 \xrightarrow{t^o} 2CO_2(1)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O(2)\\ n_{CO} = n_{CO_2} = \dfrac{4,48}{22,4} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{n_{CO_2}}{2} = 0,1(mol)\\ \Rightarrow n_{H_2} = 2n_{O_2(2)} = 2.(\dfrac{6,72}{22,4}-0,1)=0,4(mol)\\ \%V_{CO} = \dfrac{0,2}{0,2+0,4}.100\% = 33,33\%\\ \%V_{H_2} = 100\% -33,33\% = 66,67\%\)
\(\%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)
a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+27b=5,1\) (1)
Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Bảo toàn electron: \(2a+3b=0,5\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%\approx47,06\%\\\%m_{Al}=52,94\%\end{matrix}\right.\)
b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{7,3\%}=250\left(g\right)\)
\(\Rightarrow V_{HCl}=\dfrac{250}{1,2}\approx208,33\left(ml\right)\)
a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)
\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)
\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)
b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)
\(\rightarrow V_{kk}=4,48.5=11,2l\)
c. Có \(n_{Mg}=2n_{O_2}=0,2l\)
\(\rightarrow m_{Mg}=0,2.24=4,8g\)
\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)
\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)
Gọi số mol Al, Na trong a gam hỗn hợp là x, y (mol)
=> 27x + 23y = a (1)
PTHH: 4Al + 3O2 --to--> 2Al2O3
x---------------->0,5x
4Na + O2 --to--> 2Na2O
y---------------->0,5y
=> 102.0,5x + 62.0,5y = 1,64.a
=> 51x + 31y = 1,64a (2)
(1)(2) => 51x + 31y = 1,64(27x + 23y)
=> 6,72x = 6,72y
=> x = y
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27x}{27x+23y}.100\%=54\%\\\%m_{Na}=\dfrac{23y}{27x+23y}.100\%=46\%\end{matrix}\right.\)
a, Giả sử: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
⇒ 27x + 56y = 13,8 (1)
BTNT Al và Fe, có: \(\left\{{}\begin{matrix}n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}x\left(mol\right)\\n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{2}x.102+\dfrac{1}{3}y.232=21,8\left(g\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{13,8}.100\%\approx39,1\%\\\%m_{Fe}\approx60,9\%\end{matrix}\right.\)
b, BTNT O, có: \(n_{O_2}=\dfrac{3n_{Al_2O_3}+4n_{Fe_3O_4}}{2}=0,6\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)
Bạn tham khảo nhé!