a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b) \(n_{Al}=\dfrac{6,75}{27}=0,25\left(mol\right)\)

=> \(n_{H_2}=0,375\left(mol\right)\)

=> \(V_{H_2}=0,375.22,4=8,4\left(l\right)\)

c) \(n_{HCl}=0,75\left(mol\right)\)

=> \(m_{HCl}=0,75.36,5=27,375\left(g\right)\)

d) \(n_{AlCl_3}=0,25\left(mol\right)\)

=> \(m_{AlCl_3}=0,25.133,5=33,375\left(g\right)\)

\(a.pthh:2Al+6HCl--->2AlCl_3+3H_2\uparrow\)

b. Ta có: \(n_{Al}=\dfrac{6,75}{56}=\dfrac{27}{224}\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}.\dfrac{27}{224}=\dfrac{81}{448}\left(mol\right)\)

\(\Rightarrow V_{H_2}=\dfrac{81}{448}.22,4=4,05\left(lít\right)\)

c. Theo PT: \(n_{HCl}=3.\dfrac{27}{224}=\dfrac{81}{224}\left(mol\right)\)

\(\Rightarrow m_{HCl}=\dfrac{81}{224}.36,5\approx13,2\left(g\right)\)

d. Theo PT: \(n_{AlCl_3}=n_{Al}=\dfrac{27}{224}\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=\dfrac{27}{224}.133,5\approx16,09\left(g\right)\)

?????

a.    Mg +       2HCl       ZnCl2    +    H2

0,05 mol      0,1 mol                   0,05 mol

b. mMg =0,05.24 = 1,2 gam

mHClbanđầu     = mHClpu   +  mHCl dư

= 3,65 + 3,65.20% = 4,38gam

a) \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl ---> 2AlCl₃ + 3H₂O

           0,2----->1,2-------->0,4

b) \(\left\{{}\begin{matrix}m_{\text{ax}it}=m_{HCl}=1,2.36,5=43,8\left(g\right)\\m_{mu\text{ố}i}=m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\end{matrix}\right.\)

\(2Al+6HCl\to 2AlCl_3+3H_2\)

\(a) 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ b)n_{AlCl_3} = n_{Al} = \dfrac{8,1}{27} = 0,3(mol)\\ \Rightarrow m_{AlCl_3} = 0,3.133,5 = 40,05(gam)\\ c) n_{HCl} = 3n_{Al} = 0,9(mol)\\ \Rightarrow C_{M_{HCl}} = \dfrac{0,9}{0,5} = 1,8M\)

a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

            0,1      0,2                  0,1

b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,1      0,2                       0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{HCl}=0,2\cdot36,5=7,3g\)

a, Kẽm + axit clohidric → Khí hidro + muối kẽm

N_zn=0,1mol

a) có pt : Zn + 2Hcl -> ZnCl₂ + H₂

1 -> 2 -> 1 -> 1 mol

0,1-> 0,2 -> 0,1 -> o,1 mol

b) số Hcl đã dùng khi pứ là :

0,2 . 36,5= 7,3

-> số gam hcl dư là 10,95- 7,3=3,65(g)

c)m_ZnCl2=0,1 . 136=13,6g