\(\left\{{}\begin{matrix}n_{HCl}=0,55.1=0,55\left(mol\right)\\n_{H_2SO_4}=0,5.0,55=0,275\left(mol\right)\end{matrix}\right.\)
=> \(n_{H\left(trc.pư\right)}=0,55+0,275.2=1,1\left(mol\right)\)
\(n_{H_2}=\dfrac{8,736}{22,4}=0,39\left(mol\right)\)
=> \(n_{H\left(sau.pư\right)}=0,78\left(mol\right)\)
Do \(n_{H\left(trc.pư\right)}>n_{H\left(sau.pư\right)}\)
=> Axit còn dư
b)
Gọi số mol Al, Mg là a, b (mol)
=> 27a + 24b = 7,74 (1)
Giả sử công thức chung của 2 axit là HX
PTHH: 2Al + 6HX --> 2AlX3 + 3H2
a-------------------->1,5a
Mg + 2HX --> MgX2 + H2
b-------------------->b
=> 1,5a + b = 0,39 (2)
(1)(2) => a = 0,18 (mol); b = 0,12 (mol)
\(\left\{{}\begin{matrix}m_{Al}=0,18.27=4,86\left(g\right)\\m_{Mg}=0,12.24=2,88\left(g\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}n_{HCl}=0,55.1=0,55\left(mol\right)\\n_{H_2SO_4}=0,55.0,5=0,275\left(mol\right)\end{matrix}\right.\\ \rightarrow n_{H\left(trc.pư\right)}=0,55+0,275.2=1,1\left(mol\right)\\ n_{H_2}=\dfrac{8,736}{22,4}=0,39\left(mol\right)\\ \rightarrow n_{H\left(sau.pư\right)}=0,39.2=0,78\left(mol\right)\)
So sánh: \(0,78< 1,1\rightarrow\) Axit dư
Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
2Al + 6HCl ---> AlCl3 + 3H2
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
\(\rightarrow n_{H_2\left(Al\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}a=1,5a\left(mol\right)\)
Mg + 2HCl ---> MgCl2 + H2
Mg + H2SO4 ---> MgSO4 + H2
\(\rightarrow n_{H_2\left(Mg\right)}=n_{Mg}=b\left(mol\right)\)
Hệ pt \(\left\{{}\begin{matrix}27a+24b=7,74\\1,5a+b=0,39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,18\left(mol\right)\\b=0,12\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,18.27=4,86\left(g\right)\\m_{Mg}=0,12.24=2,88\left(g\right)\end{matrix}\right.\)
