\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2           0,3              0,1                 0,3

\(m_{Al}=0,2.27=5,4g\\ b.C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,45}=\dfrac{2}{3}M\\ c.2H_2+O_2\underrightarrow{t^0}2H_2O\)

0,3           0,15       0,3

\(V_{O_2}=0,15.22,4=3,36l\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ a,m_{Al}=0,2.27=5,4\left(g\right)\\ n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\\ b,C_{MddH_2SO_4}=\dfrac{0,3}{0,45}=\dfrac{2}{3}\left(M\right)\\ 2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{n_{H_2}}{2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ c,V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

nAl= 0,5(mol)

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

nHCl= 6/2 . 0,5= 1,5(mol)

=>mHCl= 1,5.36,5=54,75(mol)

=> mddHCl= (54,75.100)/18,25=300(g)

b) nH2= 3/2. 0,5=0,75(mol)

=>V(H2,đktc)=0,75.22,4=16,8(l)

c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)

mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)

=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)

Bài 2:

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ n_{Fe}=n_{H_2}=0,45\left(mol\right);n_{HCl}=2.0,45=0,9\left(mol\right)\\ a,m_{Fe}=0,45.56=25,2\left(g\right)\\ b,C_{MddHCl}=\dfrac{0,9}{0,15}=6\left(M\right)\)

Theo đè ta có : nAl = 2,7/27 = 0,1(mol)

a) PTHH :

\(4Al+3O2-^{t0}->2Al2O3\)

0,1mol...0,075mol......0,05mol

Al2O3 + 6HCl - > 2AlCl3 + 3H2O

0,05mol....0,3mol

b) Thể tích khí O2 đã dùng là :

VO2(đktc) = 0,075.22,4 = 1,6(l)

c) Khối lượng HCl đã dùng là:

mHCl = 0,3.36,5 = 10,95(g)

=> mddHCl = \(\dfrac{10,95.100}{14,6}=75\left(g\right)\)

E xem lại đề 

1)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

____0,1----->0,15

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)

2)

\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

_______0,2------------------------------>0,2

=> V_CO2 = 0,2.22,4 = 4,48(l)

3)

\(n_A=\dfrac{18,4}{M_A}\left(mol\right)\)

PTHH: 2A + Cl₂ --t^o--> 2ACl

____\(\dfrac{18,4}{M_A}\)---------->\(\dfrac{18,4}{M_A}\)

=> \(\dfrac{18,4}{M_A}\left(M_A+35,5\right)=46,8=>M_A=23\left(Na\right)\)

4)

n_HCl = 0,2.3 = 0,6(mol)

PTHH: M + 2HCl --> MCl₂ + H₂

____0,3<-----0,6

=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)

\(n_{K_2CO_3}=\dfrac{200.13,8\%}{100\%.138}=0,2(mol)\\ a,K_2CO_3+2HCl\to 2KCl+H_2O+CO_2\uparrow\\ b,n_{CO_2}=n_{K_2CO_3}=0,2(mol)\\ \Rightarrow V_{CO_2}=0,2.22,4=4,48(l)\\ c,n_{HCl}=0,4(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\)