PTHH: 2X + 2nHCl -> 2XCln + nH2
Cứ 2 mol X --> 2 mol XCln
1 mol X --> 1 mol XCln
X (g) --> X + 35,5n (g)
20 (g) --> 55,5 (g)
=> 55,5 X = 20X + 35,5n x 20
=> 35,5 X = 710n
=> X = 20n
Vì X là kim loại => n= 1;2;3
Nếu n =1 => X = 20 (Loại)
Nếu n = 2 => X = 40 (Ca)
Nếu n =3 => X = 60 (Loại)
Ta co PTHH: Ca +2HCl --> CaCl2 + H2
nCa = \(\dfrac{20}{40}\) = 0,5 mol
Cứ 1 mol Ca --> 1 mol H2
0,5 mol Ca --> 0,5 mol H2
=> \(V_{H_2}\) = 0,5 x 22,4 = 11,2 (l)
goi a la HT cua X
2X + 2aHCl \(\rightarrow\) 2XCla + aH2
pt: 2X(g) 2X+71a (g)
de: 20g 55,5g
ta co: 111X = 40X + 1420a
\(\Rightarrow\) 71X = 1420a \(\Rightarrow X=20a\)
bien luan:
* a= 1 \(\Rightarrow X=20\left(loai\right)\)
* \(a=2\Rightarrow X=40\left(lay\right)\)
* \(a=3\Rightarrow X=60\left(loai\right)\)
\(\Rightarrow\) X la Ca
\(n_{Ca}=\dfrac{20}{40}=0,5\left(mol\right)\)
Ca + 2HCl \(\rightarrow\) CaCl2 + H2
de: 0,5 \(\rightarrow\) 0,5 (mol)
\(V_{H_2}=22,4.0,5=11,2\left(l\right)\)
