Bài 42: Nồng độ dung dịch

\(C\%=\dfrac{36}{36+100}.100=26,47\%\) 
=> C

C

\(n_{NaCl}=4.0,4=1,6mol\)

\(V_{NaCl}=\dfrac{1,6}{3,2}=0,5l=500ml\)

\(V_{H_2O}=500-400=100ml\)

Mg+2Hcl->MgCl2+H2

0,3-------0,6---0,3----0,3

n Mg=0,3 mol

=>VH2=0,3.22,4=6,72l

b) CM =\(\dfrac{0,6}{0,15}=4M\)

c) C%=\(\dfrac{0,3.95}{7,2+150.1,12}100=16,2\%\)

`Mg + 2Hcl -> MgCl2 + H2`

`0,3 --- 0,6 --- 0,3 --- 0,3`

`n Mg = 0,3` `mol`

`=>` `VH2 = 0,3 . 22,4 = 6,27l`

`b)` `CM = (0,6)/(0,15) = 4M`

`c)` `C% = (0,3 . 95)/(7,2 + 150 . 1,12)100 = 16,2%`

\(n_K=\dfrac{3,51}{39}=0,09\left(mol\right)\)

PTHH: 2K + 2H₂O ---> 2KOH + H₂

        0,09--------------->0,09---->0,045

=> V_H2 = 0,045.22,4 = 1,008 (l)

m_dd = 3,51 + 46,58 - 0,045.2 = 50 (g)

=> \(C\%_{KOH}=\dfrac{0,09.56}{50}.100\%=10,08\%\)

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

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

            0,1---->0,15-------->0,05---->0,15

=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b) 

\(C\%_{dd.H_2SO_4}=\dfrac{0,15.98}{147}.100\%=10\%\)

c) 

\(m_{dd.sau.pư}=2,7+147-0,15.2=149,4\left(g\right)\)

=> \(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,05.342}{149,4}.100\%=11,446\%\)

a) \(m_{H_2SO_4}=\dfrac{150.19,6}{100}=29,4\left(g\right)\)

b) \(m_{H_2O}=150-29,4=120,6\left(g\right)\)

c) \(m_{H_2O}=50.1=50\left(g\right)\)

\(C\%=\dfrac{29,4}{150+50}.100\%=14,7\%\)

d) \(C\%=\dfrac{29,4+250.9,8\%}{250+150}.100\%=13,475\%\)

ta có :

C% NAcl=\(\dfrac{20}{20+80}.100=20\%\)

3%=\(\dfrac{mNaCl}{150}\).100

=>m Nacl=4,5g

`a) PTHH:`

`Fe + 2HCl -> FeCl_2 + H_2↑`

`0,1`                     `0,1`        `0,1`         `(mol)`

`b) n_[Fe] = [ 5,6 ] / 56= 0,1 (mol)`

 `=> V_[H_2] = 0,1 .22,4 = 2,24 (l)`

`c) C_[M_[FeCl_2]] = [ 0,1 ] / [ 0,2 ] = 0,5 (M)`