\(a)n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot0,4=0,6\left(mol\right)\\ V_{H_2}=0,6.22,4=13,44\left(l\right)\\ b)n_{HCl}=3n_{Al}=3.0,4=1,2\left(mol\right)\\ m_{HCl}=1,2.36,5=43,8\left(g\right)\\ m_{dd_{HCl}}=\dfrac{43,8}{10,95\%}\cdot100\%=400\left(g\right)\\ c)n_{AlCl_3}=n_{Al}=0,4mol\\ m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\\ m_{H_2}=0,6.2=1,2\left(g\right)\\ m_{dd_{AlCl_3}}=10,8+400-1,2=409,6\left(g\right)\\ C_{\%AlCl_3}=\dfrac{53,4}{409,6}\cdot100\%\approx13\%\)

\(n_{Al}=\dfrac{4.5}{27}=\dfrac{1}{6}\left(mol\right)\)

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

\(\dfrac{1}{6}.....0.5.......\dfrac{1}{6}.......0.25\)

\(m_{HCl}=0.5\cdot36.5=18.25\left(g\right)\)

\(m_{AlCl_3}=\dfrac{1}{6}\cdot133.5=22.25\left(g\right)\)

\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)

Yêu cầu của đề là gì vậy em ?

nAl=1/6(mol)

PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2

1/6________1/2(mol)

Ủa đề cứ thiếu thiếu , bắt tính gì làm gì em?

\(n_{Al}=\frac{10,8}{27}=0,4\left(mol\right)\) 

\(2Al+6HCl->2AlCl_3+3H_2\) (1)

theo (1) \(n_{H_2}=\frac{3}{2}n_{Al}=0,6\left(mol\right)\) 

=> \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)

1////          nAl=0,4mol   

2Al     +      6HCl -----> 2AlCl3 + 3H2

0,4mol      1,2mol         0,4mol    0,6 mol

a/ V_H2=0,6.22,4=13,44 l

b/ V=1,2/2=0,6 l

C_AlCl3=0,4/0,6=2/3 M

\(n_{Zn}=\frac{13}{65}=0,2mol\)

\(PTHH:\)

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

\(0,2\rightarrow0,4\rightarrow0,2\rightarrow0,2\left(mol\right)\)

a)\(V_{H_2}=n_{H_2}.22,4=0,2.22,4=4,48\left(l\right)\)

b)\(m_{ddspư}=m_{Zn}+m_{ddHCl}-m_{H_2}\)

                \(=13+100-0,2.1=112,8\left(gam\right)\)

\(C\%_{ddspư}=\frac{m_{ZnCl_2}.100}{m_{ddspư}}=\frac{0,2.136.100}{112,8}\approx24,1\left(\%\right)\)

PTHH: \(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)

Ta có: \(n_{HCl}=0,2\cdot0,5=0,1\left(mol\right)\) \(\Rightarrow n_{MgCl_2}=0,05mol\)

\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,05}{0,5}=0,1\left(M\right)\) (Coi như V_dd thay đổi không đáng kể)

Sửa đề: 8,4 gam Fe

\(a,n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ n_{HCl}=\dfrac{14,6.175}{36,5.100}=0,7\left(mol\right)\)

PTHH:            \(Fe+2HCl\rightarrow FeCl_2+H_2\)

ban đầu         0,15    0,7

phản ứng       0,15    0,3 

sau pư              0      0,4         0,15       0,15

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

\(b,m_{dd}=8,4+175-0,15.2=183,1\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{183,1}.100\%=10,4\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,4.36,5}{183,1}.100\%=7,97\%\end{matrix}\right.\)

a)

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(m_{HCl}=\dfrac{175.14,6}{100}=25,55\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{25,55}{35,5}=0,7\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

bđ       0,3       0,7

pư      0,3       0,6

spư      0         0,1        0,3       0,3

=> V_H2 = 0,3.22,4 = 6,72 (l)

b)

m_dd = 16,8 + 175 - 0,3.2 = 191,2 (g)

=> \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,93\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{191,2}.100\%=1,91\%\end{matrix}\right.\)

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{\dfrac{175.14,6}{100}}{36,5}=0,7\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
\(LTL:\dfrac{0,3}{1}< \dfrac{0,7}{2}\)  
\(n_{H_2}=n_{Fe}=0,3\left(mol\right)\\ V_{H_2}=0,3.22,4=6,72\left(l\right)\\ m_{\text{dd}}=16,8+175-\left(0,3.2\right)=191,2\left(g\right)\\ n_{FeCl_2}=n_{Fe}=0,3\left(mol\right)\\ C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,92\%\)
=> HCl dư