\(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,m_{ddsaup.ứ}=m_{Na}+m_{H_2O}-m_{H_2}=6,9+100-0,15.2=106,6\left(g\right)\)

Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

a. PTHH: 2Na + 2H₂O ---> 2NaOH + H₂↑

b. Ta có: \(n_{H_2O}=\dfrac{97,8}{18}=5,43\left(mol\right)\)

Ta thấy: \(\dfrac{0,1}{2}< \dfrac{5,43}{2}\)

=> H₂O dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(V_{H_2}=0,05.22,4=1,12\left(lít\right)\)

c. Ta có: \(m_{dd_{NaOH}}=2,3+97,8=100,1\left(g\right)\)

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

=> \(m_{NaOH}=0,1.40=4\left(g\right)\)

=> \(C_{\%_{NaOH}}=\dfrac{4}{100,1}.100\%=3,996\%\)

Sửa đề 2,6 -> 2,3 gam

a) $2Na + 2H_2O \to 2NaOH + H_2$

n Na = 2,3/23 = 0,1(mol)

n H2 = 1/2 n Na = 0,05(mol)

=> V H2 = 0,05.22,4 = 1,12(lít)

b) n NaOH = n Na = 2,3/23 = 0,1(mol)

=> CM NaOH = 0,1/0,2 = 0,5M

\(a) 4Na + O_2 \xrightarrow{t^o} 2Na_2O\\ b) n_{Na} = \dfrac{4,6}{23} = 0,2(mol)\\ n_{O_2} = \dfrac{1}{4}n_{Na} = 0,05(mol)\\ V_{O_2} = 0,05.22,4 = 1,12(lít)\\ c) Na_2O + H_2O \to 2NaOH\\ n_{NaOH} = n_{Na} = 0,2(mol)\\ C\%_{NaOH} = \dfrac{0,2.40}{160}.100\% = 5\%\\ d)\)

\(n_{Na\ thêm} = x(mol)\\ 2Na + 2H_2O \to 2NaOH + H_2\\ n_{NaOH} = n_{Na} = x(mol)\\ n_{H_2} =0,5x(mol)\\ \Rightarrow m_{dd} = 23x + 160 -0,5x.2 = 22x + 160(gam)\\ \Rightarrow C\% = \dfrac{0,2.40 + 40x}{22x + 160}.100\% = 5\% + 5\%\\ \Rightarrow x = \dfrac{40}{189}\\ m_{Na} = \dfrac{40}{189}.23 = 4,87(gam)\)

\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)

\(4Na+O_2\underrightarrow{^{t^0}}2Na_2O\)

\(0.2.....0.05.........0.1\)

\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.1.......................0.2\)

\(m_{NaOH}=0.2\cdot40=8\left(g\right)\)

\(C\%_{NaOH}=\dfrac{8}{160}\cdot100\%=5\%\)

Để C% tăng thêm 5% 

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

\(a...............a.......0.5a\)

\(m_{NaOH}=40a\left(g\right)\)

\(m_{dd_{NaOH}}=23a+160-0.5a\cdot2=22a+160\left(g\right)\)

\(C\%_{NaOH}=\dfrac{40a+8}{22a+160}\cdot100\%=5\%\)

\(\Rightarrow a=0\)

=> Sai đề

Bài 1 : 

a) $2Na + 2H_2O \to 2NaOH + H_2$

b) $n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol) \Rightarrow n_{Na} = 2n_{H_2} = 0,5(mol)$
$m_{Na} = 0,5.23 = 11,5(gam)$

c) $n_{NaOH} = n_{Na} = 0,5(mol)$

$C_{M_{NaOH}} = \dfrac{0,5}{0,2} = 2,5M$

$m_{H_2O} = D.V = 200.1 = 200(gam)$

$m_{dd} = 11,5 + 200 - 0,25.2 = 211(gam)$
$C\%_{NaOH} = \dfrac{0,5.40}{211}.100\% = 9,48\%$

Bài 2:

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{11,2.20\%}{22,4}=0,1\left(mol\right)\\ 4Al+3O_2\underrightarrow{^{to}}2Al_2O_3\\ Vì:\dfrac{0,1}{4}< \dfrac{0,3}{1}\Rightarrow O_2dư\\ \Rightarrow Sau.p.ứng:Al_2O_3,O_2dư,N_2\\ n_{N_2}=\dfrac{80}{20}.0,1=0,4\left(mol\right)\Rightarrow m_{N_2}=28.0,4=11,2\left(g\right)\\ n_{O_2\left(dư\right)}=0,1-\dfrac{3}{4}.0,1=0,025\left(mol\right)\\ m_{O_2\left(dư\right)}=0,025.32=0,8\left(g\right)\\ n_{Al_2O_3}=\dfrac{2}{4}.0,1=0,05\left(mol\right)\\ m_{Al_2O_3}=102.0,05=5,1\left(g\right)\)

Bài 2 : 

$n_{Al} = \dfrac{2,7}{27} = 0,1(mol) ; n_{O_2} = \dfrac{11,2.20\%}{22,4} = 0,1(mol)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$

Ta thấy : 

$n_{Al} : 4 < n_{O_2} : 3$ nên $O_2$ dư

Sau phản ứng, thu được : 

$n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,05(mol) \Rightarrow m_{Al_2O_3} = 0,05.102 = 5,1(gam)$
$n_{O_2\ pư} = \dfrac{3}{4}n_{Al} = 0,075(mol) \Rightarrow m_{O_2\ dư} = (0,1 - 0,075).32 = 0,8(gam)$
$n_{N_2} = 0,5 - 0,1 = 0,4(mol) \Rightarrow m_{N_2} = 0,4.28 = 11,2(gam)$

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a.2Na+2H_2O\rightarrow2NaOH+H_2\\ b.0,5.......0,5.........0,5..........0,25\left(mol\right)\\ b.m_{Na}=0,5.23=11,5\left(g\right)\\ c.C\%_{ddA}=C\%_{ddNaOH}=\dfrac{0,5.40}{0,5.23+200.1-0,25.2}.100\approx9,479\%\)

\(n_{HCl}=2.0,4=0,8(mol)\\ n_{Fe}=x(mol);n_{Al}=y(mol)\\ \Rightarrow 56x+27y=11(1)\\ Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow 2x+3y=0,8(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\)

\(a,\Sigma n_{H_2}=x+1,5y=0,4(mol)\\ \Rightarrow V_{H_2}=0,4.22,4=8,96(l)\\ b,m_{Fe}=0,1.56=5,6(g);m_{Al}=0,2.27=5,4(g)\\ c,m_{dd_{HCl}}=400.1,12=448(g)\\ n_{FeCl_2}=0,1(mol);n_{AlCl_3}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,1.127}{5,6+448-0,1.2}.100\%=2,8\%\\ C\%_{AlCl_3}=\dfrac{0,2.133,5}{5,4+448-0,3.2}.100\%=5,9\%\)

\(a)2Na+2H_2O\rightarrow2NaOH+H_2\\ b)n_{Na}=\dfrac{6,9}{23}=0,3mol\\2Na+2H_2O\rightarrow2NaOH+H_2\)

0,3            0,3          0,3            0,15

\(V_{H_2}=0,15.24,79=3,7455l\\ c)m_{NaOH}=0,3.40=12g\\ d)C_{\%NaOH}=\dfrac{12}{6,9+100-0,15.2}\cdot100=11,19\%\)

\(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\\ m_{HCl}=\dfrac{100.18,25}{100}=18,25\left(g\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)  
            0,125                            0,125 (mol ) 
\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\\ \)   
\(C\%=\dfrac{8,125}{8,125+18,25}.100\%=30,8\%\)
 

Bài 18:

Ta có: \(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)

\(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Xét tỉ lệ: \(\dfrac{0,125}{1}< \dfrac{0,5}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,125\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(g\right)\)

\(m_{H_2}=0,125.2=0,25\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,125\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Zn}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,25\left(mol\right)\)

Có: m _{dd sau pư} = 8,125 + 100 - 0,25 = 107,875 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,125.136}{107,875}.100\%\approx15,76\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,25.36,5}{107,875}.100\%\approx8,46\%\end{matrix}\right.\)

Bạn tham khảo nhé!

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2↑`

`0,125`  `0,25`    `0,125`    `0,125`      `(mol)`

`b)n_[Zn] = [ 8,125 ] / 65 = 0,125 (mol)`

`n_[HCl] = [ [ 18,25 ] / 100 . 100 ] / [ 36,5 ] = 0,5 (mol)`

Ta có: `[ 0,125 ] / 1 < [ 0,5 ] / 2`

    `-> Zn` hết, `HCl` dư

 `=> V_[H_2] = 0,125 . 22,4 = 2,8 (l)`

 `=> m_[H_2] = 0,125 . 2 = 0,25 (g)`

`c)m_\text{dd sau p/ứ} = 8,125 + 100 - 0,25 = 107,875 (g)`

`=> C%_[ZnCl_2] = [ 0,125 . 136 ] / [ 107,875 ] . 100 ~~ 15,76%`

`=> C%_[HCl(dư)] = [ ( 0,5 - 0,25 ) . 36,5 ] / [ 107,875 ] . 100 ~~ 8,46%`