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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
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27x}{27x+23y}.100\%=54\%\\\%m_{Na}=\dfrac{23y}{27x+23y}.100\%=46\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\Rightarrow65x+27y=2,87\left(1\right)\)
\(2Zn+O_2\underrightarrow{t^o}2ZnO\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(\Rightarrow81x+\dfrac{1}{2}y\cdot102=3,75\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,01\end{matrix}\right.\)
\(\%m_{ZnO}=\dfrac{0,04\cdot81}{3,75}\cdot100\%=86,4\%\)
Hai oxit kim loại thu được là ZnO (a mol) và Al2O3 (b mol).
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}65a+27.2b=2,87\\81a+102b=3,75\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}a=0,04\\b=0,005\end{matrix}\right.\).
Phần trăm khối lượng của kẽm oxit trong hỗn hợp sản phẩm là:
%mZnO=\(\dfrac{0,04.81}{3,75}.100\%=86,4\%\).
Theo ĐLBT KL, có: mKL + mO2 = m oxit
⇒ mO2 = 28,4 - 15,6 = 12,8 (g)
\(\Rightarrow n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)
3Fe + 2O2 --> Fe3O4 4Al + 3O2 --> 2Al2O3
x ---------------> x/3 y------------------> y/2
Theo đề bài \(\dfrac{\dfrac{x.232}{3}+\dfrac{y.102}{2}}{56x+27y}\) = \(\dfrac{283}{195}\)
Giải pt => x = 3y
=> %mFe = \(\dfrac{mFe}{mFe+mAl}.100\%\)= \(\dfrac{3y.56}{3y.56+27y}.100\%\) = 86,15%
<=> %mAl = 100 - 86,15 = 13,85%
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) \(n_{O_2}=1,5a+0,5a=0,2\left(mol\right)\)
=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{H_2}=\dfrac{0,953m}{22,4}=0,042545m\left(mol\right)\\ Đặt:n_{Mg}=x\left(mol\right);n_{Al}=y\left(mol\right);n_{Cu}=z\left(mol\right)\left(x,y,z>0\right)\\\Rightarrow \left\{{}\begin{matrix}24x+27y+64z=m\\40x+51y+80z=1,72m\\x+1,5y=0,042545m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\approx0,012845m\\y\approx0,0198m\\z\approx0,002455m\end{matrix}\right.\\ \Rightarrow\%m_{Cu}\approx\dfrac{0,002455.64m}{m}.100\%\approx15,712\%\\ \%m_{Al}\approx\dfrac{27.0,0198m}{m}.100\%\approx53,46\%\\ \%m_{Mg}\approx\dfrac{0,012845.24m}{m}.100\%\approx30,828\%\)
\(n_{O_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(a.......\dfrac{2a}{3}\)
\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)
\(b.......\dfrac{3b}{4}\)
\(n_{O_2}=\dfrac{2a}{3}+\dfrac{3b}{4}=0.25\left(mol\right)\left(1\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{H_2}=a+1.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.15,b=0.2\)
\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)
\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)
\(\%m_{Fe}=\dfrac{8.4}{8.4+5.4}\cdot100\%=60.8\%\)
\(\%m_{Al}=100-60.8=39.2\%\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(\Rightarrow\left\{{}\begin{matrix}27x+56y=22,2\\\dfrac{1}{2}x\cdot102+\dfrac{1}{3}y\cdot232=33,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)
a)\(\%m_{Al}=\dfrac{0,2\cdot27}{22,2}\cdot100\%=24,32\%\)
\(\%m_{Fe}=100\%-24,32\%=75,68\%\)
b)Theo hai pt trên:
\(\Rightarrow n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{2}{3}n_{Fe}=\dfrac{3}{4}\cdot0,2+\dfrac{2}{3}\cdot0,3=0,35mol\)
\(H=80\%\Rightarrow n_{O_2}=80\%\cdot0,35=0,28mol\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)
\(\dfrac{14}{75}\) 0,28
\(m_{KClO_3}=\dfrac{14}{75}\cdot122,5=22,87g\)
a)3 Fe+2O2--->Fe3O4
x---------2/3x-----1/3x
4Al+3O2---->2Al2O3
y---0,75y-------0,5y
n O2=15,68/22,4=0,7(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}\frac{2}{3}x+0,75y=0,7\\77,333x+51y=66,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,6\\y=0,4\end{matrix}\right.\)
m KL=0,6.56+0,4.27=44,4(g)
%m Fe=0,6.56/44,4.100%=50,45%
%m Al=100%-50,45=49,55%
b) 2KMnO4---->K2MnO4+MnO2+O2
Theo pthh
n KMnO4=2n O2=1,4(mol)
m KMnO4=1,4.158=221,2(g)
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