Gọi a,b lần lượt là số mol của Al,Mg có trong hỗn hợp ban đầu

\(n_{khí}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

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

..a............3a............a.............1,5a......(mol)

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

..b...........2b..............b...........b............(mol)

Ta có hệ phương trình: \(\left\{{}\begin{matrix}27a+24b=6,3\\1,5a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\left(mol\right)\)

\(m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\); \(m_{MgCl_2}=95.0,15=14,25\left(g\right)\)

n_HCl = 3a+2b = 2(1,5a+b) = 2.0,3 = 0,6 (mol)

\(\Rightarrow V_{HCl}=\dfrac{0,6}{0,4}.1000=1500\left(ml\right)\)\(\Rightarrow m_{ddHCl}=V_{HCl}.d=1500.1,2=1800\left(g\right)\)

\(m_{ddsauPƯ}=m_{hh}+m_{ddHCl}-m_{H_2}\)= 6,3 + 1800 - 2.0,3 = 1805,7 (g)

\(C\%_{AlCl_3}=\dfrac{13,35}{1805,7}.100\approx0,74\%\)

\(C\%_{MgCl_2}=\dfrac{14,35}{1805,7}.100\approx0,79\%\)

Mik tra loi o dua roi tick nha

\(n_{Mg}=\dfrac{3,12}{24}=0,13\left(mol\right)\); \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

....................\(Mg+S-t^o\rightarrow MgS\)

Trước PƯ:..0,13....0,1.................0.......(mol)

Trong PƯ:..0,1.......0,1...............0,1......(mol)

Sau PƯ:.....0,03.......0.................0,1.......(mol)

\(MgS+2HCl\rightarrow MgCl_2+H_2S\uparrow\)

..0,1............................................0,1.......(mol)

\(Mg_{dư}+2HCl\rightarrow MgCl_2+H_2\uparrow\)

0,03..........................................0,03.....(mol)

\(\Rightarrow\) Khí A gồm H₂S và H₂

a) \(V_{H_2S}=0,1.22,4=2,24\left(l\right)\)

\(V_{H_2}=0,03.22,4=0,672\left(l\right)\)

b) \(M_A=\dfrac{34.0,1+2.0,03}{0,1+0,03}=\dfrac{346}{13}\) (g/mol)

\(\Rightarrow\) d_A/kk = \(\dfrac{346\div13}{29}\approx0,92\)

Gọi x là sốmol HCl cần thêm (Rightarrow) (m_{HClleft(thêm ight)}=36,5xleft(g ight))

Khối lượng HCl có trong 185,4g ddHCl 10% là:

(m=185,4.dfrac{10}{100}=18,54left(g ight))

Ta có: (dfrac{36,5x+18,54}{185,4}.100=16,57)(Leftrightarrow x=0,33372left(mol ight))

(Rightarrow V_{HCl}=22,4.0,33372approx7,5left(l ight))

\(n_{KClO_3}=\dfrac{4,9}{122,5}=0,04\left(mol\right)\)

\(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\)

....0,04................................0,06...(mol)

Khối lượng oxi thu được là

\(m_{O_2}=0,06.32.\dfrac{75}{100}=1,44\left(g\right)\)