\(n_{H_2SO_4}=\dfrac{300.9,8\%}{98}=0,3\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{HCl}=0,3.2=0,6\left(mol\right)\\ n_{BaSO_4}=n_{H_2SO_4}=0,3\left(mol\right)\\ m_{ddsau}=m_{ddBaCl_2}+m_{ddH_2SO_4}-m_{BaSO_4}=200+300-233.0,3=430,1\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,6.36,5}{430,1}.100\approx5,092\%\)

Câu 1:

a) \(n_{H_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,3}=0,3\left(mol\right)\)

\(pthh:Mg+2HCl\rightarrow MgCl_2+H_2\left(1\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\)

Theo \(pthh\left(1\right);\left(2\right):n_{HCl}=2n_{H_2}=2\cdot0,3=0,6\left(mol\right)\)

\(\Rightarrow C_{M\left(HCl\right)}=\dfrac{n}{V}=\dfrac{0,6}{0,2}=3\left(M\right)\)

b) Theo \(pthh\left(1\right);\left(2\right):n_{h^2\left(Fe+Mg\right)}=n_{H_2}=0,3\left(mol\right)\)

Gọi số mol của \(Mg\) là \(x\left(mol\right)\)

Số mol của \(Fe\) là \(2x\left(mol\right)\)

\(\text{Ta có : }x+2x=0,3\\ \Leftrightarrow3x=0,3\\ \Leftrightarrow x=0,1\\ \Rightarrow n_{Mg}=0,1\left(mol\right)\\ n_{Fe}=2\cdot0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg}=n\cdot M=0,1\cdot24=2,4\left(g\right)\\ m_{Fe}=n\cdot M=0,2\cdot56=11,2\left(g\right)\\ \Rightarrow a=m_{h^2\left(Fe+Mg\right)}=2,4+11,2=13,6\left(g\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\a, 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=n_{H_2SO_4}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\ d,C\%_{ddH_2SO_4}=\dfrac{0,3.98}{200}.100\%=14,7\%\\ e,m_{ddmuoi}=5,4+200-0,3.2=204,8\left(g\right)\\ C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{34,2}{204,8}.100\%\approx16,699\%\)

\(a)2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ b)n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(0,2-\rightarrow0,3--\rightarrow0,1--\rightarrow0,3\)

\(m_{H_2}=0,3.2=0,6g\\ c)m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\\ d)C_{\%H_2SO_4}=\dfrac{0,3.98}{200}\cdot100=14,7\%\\ e)C_{\%Al_2\left(SO_4\right)_3}=\dfrac{34,2}{5,4+200-0,6}\cdot100=16,7\%\)

a) 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

b) \(m_{NaOH}=\dfrac{200.8}{100}=16\left(g\right)\)

=> \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

                0,4--->0,2--------->0,2

=> \(m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)

c) \(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)

=> \(m_{dd.H_2SO_4}=\dfrac{19,6.100}{9,8}=200\left(g\right)\)

mNaOH = 8% . 200 = 16 (g)

nNaOH = 16/40 = 0,4 (mol)

PTHH: 2NaOH + H2SO4 -> Na2SO4 + 2H2O 

Mol: 0,4 ---> 0,2 ---> 0,2 ---> 0,4

mNa2SO4 = 0,2 . 119 = 23,8 (g)

mH2SO4 = 0,2 . 98 = 19,6 (g)

mddH2SO4 = 19,6/9,8% = 200 (g)

\(m_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ m_{H_2SO_4}=200.12,25\%=24,5g\\ n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\\ LTL:0,2< 0,25=>H_2SO_4\left(d\text{ư}\right)Fe\left(h\text{ết}\right)\) 
    \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2}=0,2.22,4=4,48l\)
\(C\%=\dfrac{11,2}{24,5}.100\%=45,7\%\)

`11,2 g` j bạn ?

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)

Gọi kim loại cần tìm là X.

\(X+2HCl\rightarrow XCl_2+H_2\)

0,2                                0,2

Mà \(n_X=\dfrac{8}{M_X}=0,2\Rightarrow M_X=40\left(Ca\right)\)

Vậy kim loại cần tìm là Canxi(Ca).

a) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

          0,1-->0,2------>0,1--->0,1

=> \(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

b) m_{dd sau pư} = 5,6 + 200 - 0,1.2 = 205,4 (g)

=> \(C\%=\dfrac{12,7}{205,4}.100\%=6,18\%\)