a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl ---> 2AlCl3 + 3H2
0,2--->0,6------------------->0,3
=> VH2 = 0,3.22,4 = 6,72 (l)
b, \(m_{ddHCl}=\dfrac{0,6.36,5}{20\%}=109,5\left(g\right)\)
c, Đặt mAl = mZn = a (g)
=> \(\left\{{}\begin{matrix}n_{Al}=\dfrac{a}{27}\left(mol\right)\\n_{Zn}=\dfrac{a}{65}\left(mol\right)\end{matrix}\right.\)
PTHH:
2Al + 6HCl ---> 2AlCl3 + 3H2
\(\dfrac{a}{27}\)--------------------------->\(\dfrac{a}{18}\)
Zn + 2HCl ---> ZnCl2 + H2
\(\dfrac{a}{65}\)-------------------------->\(\dfrac{a}{65}\)
So sánh: \(\dfrac{a}{18}< \dfrac{a}{65}\)=> Al cho nhiều H2 hơn
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,6 0,3
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ m_{HCl}=20\%.0,6.36,5=4,38\left(g\right)\)
gọi nAl = nZn = a
\(pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1)
a \(\dfrac{3}{2}a\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\) (2)
a a
=> \(m_{H_2}\left(1\right)=\dfrac{3}{2}a.2=3a\left(g\right)\\ m_{H_2}\left(2\right)=a.2=2a\left(G\right)\)
=> Al sản xuất ra nhiều H2 hơn