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

PTHH :

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

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

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

\(b,m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(c,m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\)

\(m_{ddHCl}=\dfrac{43,8.100}{10}=438\left(g\right)\)

\(m_{ddAlCl_3}=10,8+438-\left(0,6.2\right)=447,6\left(g\right)\)

\(C\%=\dfrac{53,8}{447,6}.100\%\approx12,02\%\)

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

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2\cdot56=11,2\left(g\right)\\m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\m_{ddHCl}=\dfrac{0,4\cdot36,5}{10\%}=146\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Fe}+m_{ddHCl}-m_{H_2}=156,8\left(g\right)\) \(\Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{156,8}\cdot100\%\approx16,2\%\)

mH2SO4=9,8%.300=29,4(g)

=> nH2SO4=0,3(mol)

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

0,2<-------------0,3----------->0,1------------->0,3(mol)

b) a=mAl=0,2.27=5,4(g)

c) V(H2,đktc)=0,3.22,4=6,72(l)

d) mAl2(SO4)3=0,1.342=34,2(g)

e) mddAl2(SO4)3=mAl+mddH2SO4- mH2= 5,4+300-0,3.2= 304,8(g)

=> C%ddAl2(SO4)3=(34,2/304,8).100=11,22%

a)         2Al+ 3H₂SO₄→ Al₂(SO₄)₃+ 3H₂

(mol)     0,2       0,3          0,1           0,3                                                    

b) m H₂SO₄= 300. 9,8%= 29,4(g)

n H₂SO₄= \(\dfrac{m}{M}=\dfrac{29,4}{98}=0,3\)(mol)

c) V H₂= n.22,4= 0,3.22,4= 6,72(lít)

m H₂= n.m= 0,3.2= 0,6(g)

d) m Al₂(SO₄)₃= n.M= 0,1.342= 34,2(g)

e) m_Al= n.M= 0,2.27= 5,4(g)

mdd_{sau phản ứng}= m_Al+ mdd H₂SO₄- m H₂

                          = 5,4+300-0,6= 304,8(g)

=> C%dd_{sau phản ứng}= \(\dfrac{34,2}{304,8}.100\%=11,22\%\)

asd

\(m_{ct}=\dfrac{7,3.300}{100}=21,9\left(g\right)\)

\(n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2|\)

        2          6              2            3

       0,2      0,6            0,2         0,3

a) \(n_{Al}=\dfrac{0,6.2}{6}=0,2\left(mol\right)\)

⇒ \(m_{Al}=0,2.27=5,4\left(g\right)\)

b) \(n_{AlCl3}=\dfrac{0,6.2}{6}=0,2\left(mol\right)\)

⇒ \(m_{AlCl3}=0,2.133,5=26,7\left(g\right)\)

c) \(n_{H2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

d) \(m_{ddspu}=5,4+300-\left(0,3.2\right)=304,8\left(g\right)\)

\(C_{AlCl3}=\dfrac{26,7.100}{304,8}=8,76\)⁰/₀

 Chúc bạn học tốt

a)

$2Al +3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)

$n_{H_2} = n_{H_2SO_4} = \dfrac{300.9,8\%}{98} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$

c)

$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,1(mol)$
$m_{Al_2(SO_4)_3} = 0,1.342 = 34,2(gam)$

d)

$n_{Al} = \dfrac{2}{3}n_{H_2SO_4} = 0,2(mol)$
$m_{dd} = 0,2.27 + 300 - 0,3.2 = 304,8(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{34,2}{304,8}.100\% = 11,22\%$

nH2SO4=0,3(mol)

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

a) 0,2_______0,3______0,1______0,3(mol)

b) V(H2,đktc)=0,3.22,4=6,72(l)

c) a=mAl=0,2.27=5,4(g)

=>a=5,4(g)

d)  mAl2(SO4)3=342.0,1=34,2(g)

e) mddAl2(SO4)3= 5,4+ 300 - 0,3.2= 304,8(g)

=>C%ddAl2(SO4)3= (34,2/304,8).100=11,22%

a) Phương trình hóa học:

              2Al+ 3H₂SO₄→ Al₂(SO₄)₃+ 3H₂

( mol)      0,2       0,3            0,1          0,3                                        

b) m H₂SO₄= \(\dfrac{9,8\%.300}{100\%}=29,4\)(gam)

→n H₂SO₄= \(\dfrac{m}{M}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

=> m_Al= \(n.m=0,2.27=5,4\left(gam\right)\)

c) V H₂= n.22,4= 0,3.22,4= 6,72( lít)

m H₂= n.M= 0,3.2= 0,6(gam)

d) m Al₂(SO₄)₃= n.M= 0,1.342= 34,2(gam)

e) m_{dd sau phản ứng}= m_Al+ m_ddH₂SO₄- m H₂

                                    = 5,4  + 300- 0,6= 304,8(gam)

=> C%_{dd sau phản ứng}=\(\dfrac{34,2}{304,8}.100\%=11,22\%\)

a) K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

b) \(n_{K_2CO_3}=\dfrac{13,8}{138}=0,1\left(mol\right)\)

PTHH: K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

______0,1----->0,2------>0,2--->0,1

=> V_CO2 = 0,1.22,4 = 2,24 (l)

c) m_HCl = 0,2.36,5 = 7,3 (g)

\(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)

d) m_KCl = 0,2.74,5 = 14,9 (g)

m_{dd sau pư} = 13,8 + 100 - 0,1.44 = 109,4 (g)

=> \(C\%\left(KCl\right)=\dfrac{14,9}{109,4}.100\%=13,62\%\)

n_Al = 5.4 / 27 = 0.2 (mol)

2Al + 6HCl => 2AlCl₃ + 3H₂

0.2......0.6............0.2.......0.3

a) V_H2 = 0.3 * 22.4 = 6.72 (l) 

b) m_AlCl3 = 0.2 * 133.5 = 26.7 (g) 

c) VddHCl = 0.6 / 1.5 = 0.4 (l) 

d) C_MAlCl3 = 0.2 / 0.4 = 0.5 (M) 

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

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

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)

500ml = 0,5l

\(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)

a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

              1         2           1          1

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

b) \(n_{Fe}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

⇒ \(m_{Fe}=0,05.56=2,8\left(g\right)\)

c) \(n_{H2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)

d) \(n_{FeCl2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

⇒ \(m_{FeCl2}=0,05.127=6,35\left(g\right)\)

 Chúc bạn học tốt

cíuuu em mn ớiii. maii em thii:(

nHCl = 0,5.0,2= 0,1(mol)

a)  Fe    +   2HCl → FeCl2+ H2

     0,05 ←  0,1   → 0,05   →0,05

b)  mFe = 0,05.56=2,8(gam)

c) V_H2= 0,05 .22,4 =1,12 (lít)

d)  mFeCl2 = 0,05.(56+35,5.2)=6,35(gam)