a) PTHH : 2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow n_{H_2}=n_{Al}\cdot\dfrac{3}{2}=0,2\cdot\dfrac{3}{2}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=n\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\)
b) \(n_{HCl}=n_{Al}\cdot\dfrac{6}{2}=0,2\cdot\dfrac{6}{2}=0,6\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{n}{C_M}=\dfrac{0,6}{1,5}=0,4\left(l\right)\)
a, PTHH: 2Al+6HCl ---> 2AlCl3+3H2
nAl=5,4/27=0,2 mol
Theo PTHH ta có: nH2= 3/2. nAl= 3/2. 0,2= 0,3 mol
=> VH2=0,3.22.4= 6,72 l
b, Theo pthh ta có: nHCl= 6/2. nAl= 6/2. 0,2= 0,6 mol
=> VddHCl= n/CM= 0,6/1,5= 0,4 (l)
PTHH: 2Al + 6HCl \(\rightarrow\) 2AlCl3 + H2
a) So mol Al tham gia phan ung la
nAl = \(\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PTHH ta có
\(n_{H_2}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}\cdot0,2=0,1\left(mol\right)\)
Thể tích khí Hiđro cần dùng ở đktc là
\(V_{H_2}=n_{H_2}\cdot22,4=0,1\cdot22.4=2.24\left(l\right)\)
Vậy thể tích hiđro cần dùng là 2,24 l
