\(n_{K2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)

Pt : \(K_2O+H_2O\rightarrow2KOH|\)

         1           1              2

        0,1                        0,2

a) \(n_{KOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

200ml = 0,2l

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

b) Pt : \(KOH+HCl\rightarrow KCl+H_2O|\)

              1            1           1            1

            0,1          0,1

\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(m_{ddHCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)

c) \(CO_2+2KOH\rightarrow K_2CO_3+H_2O|\)

        1          2               1              1

      0,05      0,1

\(n_{CO2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

\(V_{CO2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)

 Chúc bạn học tốt

 \(PTHH:4Al+6HCl\rightarrow2Al_2Cl_3+3H_2\uparrow\)

\(n_{Al}=\frac{3,78}{27}=0,14\left(mol\right)\)

\(\Rightarrow n_{H_2}=\frac{3}{4}n_{Al}=0,105\left(mol\right)\)

\(V_{H_2}=0,105.22,4=2,352\left(l\right)\)

\(n_{HCl}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,14=0,21\left(mol\right)\)

\(C_{M_{ddHCl}}=\frac{0,21}{0,2}=1,05\left(M\right)\)

\(n_{Al_2Cl_3}=\frac{1}{2}n_{Al}=\frac{1}{2}.0,14=0,07\left(mol\right)\)

\(m_{Al_2Cl_3}=0,07.160,5=11,235\left(g\right)\)

a) $CaSO_3 + 2HCl \to CaCl_2 + SO_2 + H_2O$
b)

$n_{SO_2} = n_{CaSO_3} = \dfrac{12}{120} = 0,1(mol)$
$m_{SO_2} = 0,1.64 = 6,4(gam)$

c)

$n_{HCl} = 2n_{SO_2} = 0,2(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$

d)

$m_{dd\ sau\ pư} = m_{CaSO_3} + m_{dd\ HCl} - m_{SO_2} = 12 + 50 - 6,4 = 55,6(gam)$

$C\%_{CaCl_2} = \dfrac{0,1.111}{55,6}.100\% = 19,96\%$

Ta có: \(n_{CaSO_3}=\dfrac{12}{120}=0,1\left(mol\right)\)

a. PTHH: CaSO₃ + 2HCl ---> CaCl₂ + H₂O + SO₂

b. Theo PT: \(n_{SO_2}=n_{CaSO_3}=0,1\left(mol\right)\)

=> \(m_{SO_2}=0,1.64=6,4\left(g\right)\)

c. Theo PT: \(n_{HCl}=2.n_{CaSO_3}=2.0,1=0,2\left(mol\right)\)

=> \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{7,3}{m_{dd_{HCl}}}.100\%=14,6\%\)
=> \(m_{dd_{HCl}}=50\left(g\right)\)

d. Ta có: \(m_{dd_{CaCl_2}}=12+50-0,1.64=55,6\left(g\right)\)

Theo PT: \(n_{CaCl_2}=n_{SO_2}=0,1\left(mol\right)\)

=> \(m_{CaCl_2}=0,1.111=11,1\left(g\right)\)

=> \(C_{\%_{CaCl_2}}=\dfrac{11,1}{55,6}.100\%=19,96\%\)

A) có 2 pthh

Na2o + h2o ----> 2Naoh

2Na +2 h2o ------> 2naoh + h2

N khí. H2 = 0,56/22,4 =0,025 (mol)

Gọi x và y lần lượt là số mol của bà và na2o

Viết lại pt

2Na +2 h2o----> 2 naoh + h2

X mol.                                  X/2 moll

Na2o + h2o-----> 2naoh

Xin lỗi bài này có gif đó sai sai xin bí tay

nAl= 0,5(mol)

a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2

nHCl= 6/2 . 0,5= 1,5(mol)

=>mHCl= 1,5.36,5=54,75(mol)

=> mddHCl= (54,75.100)/18,25=300(g)

b) nH2= 3/2. 0,5=0,75(mol)

=>V(H2,đktc)=0,75.22,4=16,8(l)

c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)

mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)

=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH :

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

0,2     0,4            0,2        0,2 

\(a,V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(b,m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{ddHCl}=\dfrac{14,6.100}{10}=146\left(g\right)\)

\(c,m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(m_{ddMgCl_2}=4,8+146-\left(0,2.2\right)=150,4\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{19}{150,4}.100\%\approx12,63\%\)

2. 

\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

\(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)

0,2                     0,2         0,1 

\(m_{KOH}=0,2.56=11,2\left(g\right)\)

\(m_{ddKOH}=7,8+100-\left(0,1.2\right)=107,6\left(g\right)\)

\(C\%=\dfrac{11,2}{107,6}.100\%\approx10,4\%\)

a) $Zn + 2HCl \to ZnCl_2 + H_2$

b) Theo PTHH : $n_{H_2} = n_{Zn} = \dfrac{16,25}{65} = 0,25(mol)$
$\Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)$

c) $n_{HCl} = 2n_{Zn} = 0,5(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,5}{0,5} = 1M$

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

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

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=11.2+200-0.2\cdot2=210.8\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{25.4}{210.8}\cdot100\%=12.05\%\)

Giúp với mai mình thi rồi!