a. PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ 0,2mol:0,6mol\rightarrow0,2mol:0,3mol\)
b. \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
Ta có tỉ lệ: \(\dfrac{0,2}{2}=0,1< \dfrac{0,8}{6}=\dfrac{2}{15}\)
Vậy Al PƯ hết, HCl PƯ dư.
\(m_{HCldu}=m_{HClbandau}-m_{HClpu}\)
\(\Leftrightarrow m_{HCldu}=29,2-36,5.0,6=29,2-21,9=7,3\left(g\right)\)
a)
PTHH: 2Al + 6HCl \(\rightarrow\) 2AlCl\(_3\) + 3H\(_2\)
b)
Ta có:
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
PTHH: 2Al + 6HCl \(\rightarrow\) 2AlCl\(_3\) + 3H\(_2\)
(mol)PT: 2 6
(mol)đề: 0,2 0,8
Ta có tỉ lệ:
\(\dfrac{0,2}{2}< \dfrac{0,8}{6}\) \(\Rightarrow\) Sau PƯ Al hết, HCl dư
Ta có:
\(n_{HCldu}=n_{HCl}-n_{HClpu}=0,8-0,6=0,2\left(mol\right)\)
=> \(m_{HCldu}=n_{HCldu}.M_{HCl}=0,2.\left(1+35,5\right)=7,3\left(g\right)\)
\(\Rightarrow\)\(m_{HCldu}=m_{HClbanđầu}-m_{HClpu}=29,2-7,3=21,9\left(g\right)\)
a) 2Al + 6HCl ----> 2AlCl3 + 3H2
b)
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{m}{M}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
Ta có :
Theo PTHH : 3nAl = nHCl
Theo giả thiết : 3nAl < nHCl
.................(0,2 . 3 = 0,6 < 0,8 )
=> HCl dư
=> \(n_{HCl}dư=0,8-\left(0,2\cdot3\right)=0,2\left(mol\right)\)
=> mHCl dư = n . M =0,2 . 36,5 = 7,3 (g)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
a) PTHH: 2Al + 6HCl → 2AlCl3 + 3H2↑
Ban đầu: 0,2........0,8......................................(mol)
Phản ứng: 0,2........0,6......................................(mol)
Sau phản ứng: 0...........0,2...→....0,2.............0,3...(mol)
b) Vậy sau phản ứng HCl dư
\(m_{HCl}dư=0,2\times36,5=7,3\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
a. PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b. Theo PT và đề bài ta có tỉ lệ:
\(\dfrac{0,2}{2}< \dfrac{0,8}{6}\Rightarrow HCl_{dư}\); Al hết. Nên ta tính theo \(n_{Al}\)
Theo PT ta có: \(n_{HCl\left(pư\right)}=\dfrac{0,2.6}{2}=0,6\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,6=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)
nAl = \(\dfrac{5,4}{27}=0,2mol\)
nHCl = \(\dfrac{29,2}{36,5}=0,8mol\)
a) PTHH: 2Al + 6HCl → 2AlCl3 + 3H2\(\uparrow\)
Ban đầu: 0,2........0,8......................................(mol)
Phản ứng: 0,2........0,6......................................(mol)
Sau phản ứng: 0...........0,2...→....0,2.............0,3...(mol)
b) Vậy sau phản ứng HCl dư
mHCl dư = 0,2 . 36,5 = 7,3 g
