Gọi x,y(mol) là số mol Al,Mg.
PTHH: 2Al +6HCl---> 2AlCl3+ 3H2
............x........................................3/2.x(mol)
PTHH: Mg+ 2HCl----> MgCl2+H2
.............y..........................................y(mol)
Ta có:\(\left\{{}\begin{matrix}27x+24y=1,95\\\frac{3}{2}x+y=\frac{2,24}{22,4}=0,1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{20}\\y=\frac{1}{40}\end{matrix}\right.\)(TM)
K/l Al:
.................\(m_{Al}=\frac{1}{20}.27=\frac{27}{20}\left(g\right)\)
mMg=24/40(G)
#Walker
Gọi x,y lần lượt là số mol của Al,Mg
PTHH: 2Al+6HCl->2AlCl\(_3\)+3H\(_2\)
mol:......2.......6.........2..............3
mol:....x.........3x...........x.........1,5x
PTHH:Mg+2HCl->MgCl\(_2\)+H2
mol:.......1.....2..........1............1
mol:.........y.......2y........y..........y
ta có:27x+24y=1,95
(1,5x+y)22,4=2,24
=>x=0,05(mol),y=0,025(mol)
m\(_{Al}\)=0,05*27=1,35(g)
m\(_{Mg}\)=1,95-1,35=0,6(g)
\(n_{Al}=x;n_{Mg}=y\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ hpt:\left\{{}\begin{matrix}27x+24y=1,95\\1,5x+y=\frac{2,24}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,025\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}m_{Al}=0,05.27=1,35\left(g\right)\\m_{Mg}=0,025.24=0,6\left(g\right)\end{matrix}\right.\)
Gọi số mol Al là a (a>0)
Số mol Mg là b( b>0)
pthh : 2Al + 6HCl ---> 2AlCl3 + 3H2
a 3a/2
Mg + 2HCl -----> MgCl2 + H2
b b
Theo bài ra ta có pt : 27a + 24 b = 1,95
3/2.a + b = 2,24 + 22,4
<=> a = 0,05
b = 0,025
=> %mAl = (0,05.27):1,95 . 100% ≈ 69,23%
=> % m Mg = 100% - 69,23% =30,77%
