\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)

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

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

b)\(m_{ZnCl_2}=0,05\cdot136=6,8g\)

c)\(V_{H_2}=0,05\cdot22,4=1,12l\)

a) Nhôm + Axit clohidric --> Nhôm clorua + khí hidro

b) 2Al+ 6HCl --> 2AlCl₃ + 3H₂

c) Theo ĐLBTKL: m_Al + m_HCl = m_AlCl3 + m_H2

=> m_AlCl3 = 5,4+21,9-0,6 = 26,7(g)

d) Số nguyên tử nhôm : số phân tử HCl = 2:6 

Số phân tử AlCl₃ : số phân tử H₂ = 2:3

\(a.Al+HCl\rightarrow AlCl_3+H_2\\ b.Al+2HCl\rightarrow AlCl_3+H_2\)

\(c.m_{Al}+m_{HCl}=m_{AlCl_3}+m_{H_2}\\ \Rightarrow m_{AlCl_3}=m_{Al}+m_{HCl}-m_{H_2}=5,4+21,9-0,6=26,7\left(g\right)\)

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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

b, Ta có: \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

Bạn tham khảo nhé!

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,2mol\\n_{H_2}=0,3mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)

B là khí hi đro nha

\(n_{Al}=\dfrac{13.5}{27}=0.5\left(mol\right)\)

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

\(0.5........................0.5......0.75\)

\(a.\)

\(m_{AlCl_3}=0.5\cdot133.5=66.75\left(g\right)\)

\(b.\)

\(V_{H_2}=0.75\cdot22.4=16.8\left(l\right)\)

\(c.\)

\(n_{FeO}=\dfrac{64.8}{72}=0.9\left(mol\right)\)

\(FeO+H_2\underrightarrow{t^0}Fe+H_2O\)

\(0.75.....0.75...0.75\)

\(\Rightarrow FeOdư\)

\(m_{Fe}=0.75\cdot56=42\left(g\right)\)

a)mol Al= 0.5

pthh:     2Al + 6HCl -----> 2AlCl\(_3\) + 3H\(_2\)

0.5mol Al= 0,5 mol AlCl\(_3\) ---> m\(_{AlCl}\)\(_3\)=0,5*(27+35,5*3)=66.75

c) pthh:   FeO + H\(_2\) --\(^{to}\)-> Fe + H\(_2\)O

mol\(_{FeO}\)=64,8/(56+16)=0,9

mol\(_{H_{ }2}\)=0.75

mol\(h2\)<mol\(_{FeO}\)---->H2 hết,   FeO dư

m\(_{Fe}\)=0,75*56=42(g)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

______0,2-->0,6------->0,2---->0,3

=> m_AlCl3 = 0,2.133,5 = 26,7(g)

b) V = 0,3.22,4 = 6,72(l)

c) Số phân tử HCl = 0,6.6.10²³ = 3,6.10²³

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{FeCl_2} = n_{Fe} =\dfrac{11,2}{56} = 0,2(mol)\\ m_{FeCl_2} = 0,2.127 = 25,4(gam)\\ c) n_{H_2} = n_{Fe} = 0,2(mol)\Rightarrow V_{H_2} = 0,2.22,4 = 4,48(lít)\\ d) n_{HCl} = 2n_{Fe} = 0,4(mol)\\ C\%_{HCl} = \dfrac{0,4.36,5}{300}.100\% = 4,867\%\)

a) $2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
b) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
Theo PTHH : $n_{Cl_2} = \dfrac{3}{2}n_{Al} =  0,3(mol)$
$\Rightarrow V_{Cl_2} = 0,3.24,79 = 7,437(lít)$

c) $n_{AlCl_3} = n_{Al} = 0,2(mol)$
$\Rightarrow m_{AlCl_3} = 0,2.133,5 = 26,7(gam)$