a)
Gọi số mol Al, Mg trong mỗi phần là a, b (mol)
Phần 1:
\(n_{H_2}=\dfrac{5,152}{22,4}=0,23\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PTHH: \(n_{H_2}=1,5a+b=\) 0,23 (1)
Phần 2:
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(2Mg+O_2\underrightarrow{t^o}2MgO\)
Theo PTHH: \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,5a\left(mol\right)\\n_{MgO}=b\left(mol\right)\end{matrix}\right.\)
=> 102.0,5a + 40b = 8,3
=> 51a + 40b = 8,3 (2)
(1)(2) => a = 0,1 (mol); b = 0,08 (mol)
=> m = 2.(0,1.27 + 0,08.24) = 9,24 (g)
b) Theo PTHH: \(n_{O_2}=0,75a+0,5b\) = 0,115 (mol)
=> \(V_{O_2}=0,115.22,4=2,576\left(l\right)\)
