Câu 9 : 

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

a) Pt : \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

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

→ \(V_{H2\left(dtkc\right)}=0,2.22,4=4,48\left(l\right)\)

b) \(V_{ddCH3COOH}=\dfrac{0,4}{2}=0,2\left(l\right)\)

c) \(m_{\left(CH3COO\right)2Mg}=0,2.101=20,2\left(g\right)\)

d) Pt : \(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)

                0,4                0,4

\(C_{MddKOH}=\dfrac{0,4}{0,2}=2\left(M\right)\)

 Chúc bạn học tốt

\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1       0,2            0,1         0,1

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

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

\(V_{H_2}=\left(0,1+0,2\right).22,4=6,72l\\ b)V_{ddHCl}=\dfrac{0,2+0,4}{2}=0,3l\\ c)m_{muối}=0,1.127+95.0,2=31,7g\)

a)

$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2o$

b)

Theo PTHH : $n_{Mg} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$m_{Mg} = 0,2.24 = 4,8(gam)$

$m_{MgO} = m_{hh} - m_{Mg} = 12,8 - 4,8 = 8(gam)$

c)

$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$
$n_{HCl} = 2n_{Mg} + 2n_{MgO} = 0,8(mol)$

$m_{dd\ HCl} = \dfrac{0,8.36,5}{14,6\%} = 200(gam)$

a)

$CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O$

b)

n NaHCO3 = n CH3COOH = 100.12%/60 = 0,2(mol)

m dd NaHCO3 = 0,2.84/8% = 210(gam)

c)

n CO2 = n CH3COOH = 0,2(mol)

=> V CO2 = 0,2.22,4 = 4,48(lít)

d)

m dd = m dd CH3COOH + m dd NaHCO3 - m CO2 = 100  + 210 - 0,2.44 = 301,2(gam)

C% CH3COONa = 0,2.82/301,2   .100% = 5,44%

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

a, Ta có: 65n_Zn + 27n_Al = 11,9 (1)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)

⇒ m_Zn = 0,1.65 = 6,5 (g)

m_Al = 0,2.27 = 5,4 (g)

b, Theo PT: n_ZnCl2 = n_Zn = 0,1 (mol)

n_AlCl3 = n_Al = 0,2 (mol)

⇒ m _muối = 0,1.136 + 0,2.133,5 = 40,3 (g)

c, Theo PT: n_HCl = 2n_H2 = 0,8 (mol)

\(\Rightarrow m_{ddHCl}=\dfrac{0,8.36,5}{10\%}=292\left(g\right)\)

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

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

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