1
\(n_O=\dfrac{16-11,2}{16}=0,3\left(mol\right)\)
=> \(n_{H_2O}=0,3\left(mol\right)\)
=> \(n_{H_2}=0,3\left(mol\right)\)
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
2
\(n_{HCl}=0,4.1=0,4\left(mol\right)\)
PTHH: X + 2HCl --> XCl2 + H2
0,2<--0,4
=> \(M_X=\dfrac{13}{0,2}=65\left(g/mol\right)\)
=> X là Zn
1.
ta có :
m O=16-11,2=4,8g
n O=\(\dfrac{4,8}{16}\)=0,3 mol
=>n H2=n H2O=n O=0,3 mol
=>VH2=0,3.22,4=6,72l
2
X+2HCl->XCl2+H2
0,2---0,4 mol
n HCl=0,4.1=0,4 mol
ta có :\(\dfrac{13}{X}=0,2\)
=>MX=65 đvC
=>X là kẽm (Zn)