\(a.n_{Fe}=\dfrac{8,4}{56}=0,15mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=n_{FeCl_2}=0,15mol\\ V_{H_2,đktc}=0,15.22,4=3,36l\\ V_{H_2,đkc}=0,15.24,79=3,7185l\\ b.n_{HCl}=0,15.2=0,3mol\\ C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5M\\ c.m_{FeCl_2}=0,15.127=19,05g\)

nMg = 6/24 = 0,25 (mol)

PTHH: Mg + 2HCl -> MgCl2 + H2

nH2 = 0,25 (mol(

VH2 = 0,25 . 24,79 = 6,1975 (l)

CuO + H2 -> (t°) Cu + H2O

nCu = 0,25 (mol)

mCu = 0,25 . 64 = 16 (g)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH :

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

0,1        0,1             0,1             0,1 

\(a,m_{MgSO_4}=0,1.120=12\left(g\right)\)

\(b,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(c,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{10}=98\left(g\right)\)

\(a.2Al+6HCl->2AlCl_3+3H_2\\ b.m_{AlCl_3}=\dfrac{1}{3}.0,1.0,6.133,5=2,67g\\ c.V_{H_2}=\dfrac{1}{2}.0,06.24,79=0,7437\left(L\right)\\ d.a=\dfrac{1}{3}.0,1.0,6.27=0,54g\)

Bài 14:

Ta có: \(n_{BaCO_3}=\dfrac{39,4}{197}=0,2\left(mol\right)\)

PT: \(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\)

a, \(n_{CO_2}=n_{BaCO_3}=0,2\left(mol\right)\Rightarrow V_{CO_2}=0,2.24,79=4,958\left(l\right)\)

b, Sửa đề: tính khối lượng dung dịch HCl → tính nồng độ % dd HCl.

 \(n_{HCl}=2n_{BaCO_3}=0,4\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{100}.100\%=14,6\%\)

c, \(n_{BaCl_2}=n_{BaCO_3}=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 39,4 + 100 - 0,2.44 = 130,6 (g)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,2.208}{130,6}.100\%\approx31,85\%\)

Bài 12:

Ta có: \(n_{MgCO_3}=\dfrac{25,2}{84}=0,3\left(mol\right)\)

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

a, Theo PT: \(n_{CO_2}=n_{MgCO_3}=0,3\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,3.24,79=7,437\left(l\right)\)

b, Ta có: m _{dd sau pư} = 25,2 + 200 - 0,3.44 = 212 (g)

Theo PT: \(n_{MgCl_2}=n_{MgCO_3}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,3.95}{212}.100\%\approx13,44\%\)

Bài 13:

Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

1. \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\) \(\Rightarrow V_{CO_2}=0,1.24,79=2,479\left(l\right)\)

2. \(n_{HCl}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{7,3\%}=100\left(g\right)\)

3. Ta có: m _{dd sau pư} = 10 + 100 - 0,1.44 = 105,6 (g)

Theo PT: \(n_{CaCl_2}=n_{CaCO_3}=0,1\left(mol\right)\)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,1.111}{105,6}.100\%\approx10,51\%\)

\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,V_{H_2\left(25\text{đ}\text{ộ}C,1bar\right)}=0,2.24,79=4,958\left(l\right)\\ c,n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{\text{dd}HCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ m_{\text{dd}A}=m_{Mg}+m_{\text{dd}HCl}-m_{H_2}=4,8+146-0,2.2=150,4\left(g\right)\\ d,C\%_{\text{dd}MgCl_2}=\dfrac{95.0,2}{150,4}.100\approx12,633\%\)

a) 2Al+6HCl2AlCl3+3H2

b)

nAl==0,4(mol)

nAlCl3=nAl=0,4(mol)

mAlCl3=0,4.133,5=53,4(g)

c)

nH2=nAl=0,6(mol)

VH2=22,4.0,6=13,44(l)

d) n HCl=0,4.6\2=1,2 mol

=>Cm HCl=1,2\0,1=12M

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

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

            2           6              2            3

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

b) \(n_{H2}=\dfrac{0,4.3}{2}=0,6\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,6.24,79=14,874\left(l\right)\)

c)  \(n_{AlCl3}=\dfrac{0,6.2}{3}=0,4\left(mol\right)\)

⇒ \(m_{AlCl3}=0,4.133,5=53,4\left(g\right)\)

d) \(n_{HCl}=\dfrac{0,4.6}{2}=1,2\left(mol\right)\)

100ml = 0,1l

\(C_{M_{ddHCl}}=\dfrac{1,2}{0,1}=12\left(M\right)\)

 Chúc bạn học tốt

\(a,PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ b,n_{FeCl_2}=n_{H_2}=n_{Fe}=0,4\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ c,m_{FeCl_2}=127.0,4=50,8\left(g\right)\)