a) 2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

b) \(n_{CH_3COOH}=\dfrac{200.12\%}{60}=0,4\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

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

=> \(m_{Na_2CO_3}=0,2.106=21,2\left(g\right)\)

=> \(m_{dd.Na_2CO_3}=\dfrac{21,2.100}{50}=42,4\left(g\right)\)

c) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

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

PT: \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)

a, Theo PT: \(n_{CH_3COOH}=2n_{Fe}=0,2\left(mol\right)\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{10\%}=120\left(g\right)\)

\(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 5,6 + 120 - 0,1.2 = 125,4 (g)

\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{0,1.174}{125,4}.100\%\approx13,88\%\)

cái gí tu

\(a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\b,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\Rightarrow n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\left(mol\right);n_{HCl}=2.0,1=0,2\left(mol\right)\\ b,m_{ddHCl}=\dfrac{0,2.36,5.100}{20}=36,5\left(g\right)\\ c,m_{ddsau}=2,4+36,5-0,1.2=38,7\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,1.95}{38,7}.100\approx24,548\%\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{20\%}=36,5\left(g\right)\)

c, \(n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 2,4 + 36,5 - 0,1.2 = 38,7 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,1.95}{38,7}.100\%\approx24,55\%\)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeSO_4}=0,2.152=30,4\left(g\right)\)

c, \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

d, \(n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

\(a,Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{Mg}=\dfrac{12}{24}=0,5mol\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,5mol\\ V_{H_2}=0,5.24,79=12,395l\\ c,n_{HCl}=0,5.2=1mol\\ C_{M_{HCl}}=\dfrac{1}{0,1}=10M\\ d,m_{MgCl_2}=0,5.95=47,5g\)

`#3107.101107`

`a,`

PTHH: \(\text{Mg}+\text{2HCl}\rightarrow\text{ MgCl}_2+\text{H}_2\)

`b,` 

n của Mg trong pứ là:

\(\text{n}_{\text{Mg}}=\dfrac{\text{m}_{\text{Mg}}}{\text{M}_{\text{Mg}}}=\dfrac{12}{24}=\dfrac{1}{2}=0,5\left(\text{mol}\right)\)

Theo pt: `1` mol Mg thu được `1` mol H₂

`=>`\(\text{n}_{\text{H}_2}=0,5\left(\text{mol}\right)\)

Thể tích của H₂ thu được sau pứ là:

\(\text{V}_{\text{H}_2}=\text{n}_{\text{H}_2}\cdot24,79=0,5\cdot24,79=12,395\left(l\right)\)

c.

Đổi `100` ml = `0,1` l

Theo PT: `1` mol Mg tác dụng với `2` mol HCl được pứ trên

`=>`\(\text{n}_{\text{HCl}}=2\cdot0,5=1\left(\text{mol}\right)\)

Nồng độ `%` của HCl là:

\(\text{C}_{\text{M}}=\dfrac{\text{n}_{HCl}}{\text{V}_{\text{HCl}}}=\dfrac{1}{0,1}=10\left(\text{M}\right)\)

d.

Theo PT: `1` mol Mg thu được `1` mol MgCl₂

`=>`\(\text{n}_{\text{MgCl}_2}=0,5\text{ (mol)}\)

m của MgCl₂ thu được sau pứ là:

\(\text{m}_{\text{MgCl}_2}=\text{n}_{\text{MgCl}_2}\cdot\text{M}_{\text{MgCl}_2}=0,5\cdot95=47,5\left(g\right).\)

`a)PTHH:`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`0,2`     `0,6`                               `0,3`       `(mol)`

`n_[Al]=[5,4]/27=0,2(mol)`

`b)V_[H_2]=0,3.22,4=6,72(l)`

`c)m_[dd HCl]=[0,6.36,5]/10 . 100 =219(g)`

n_Al  = \(\dfrac{5,4}{27}\) = 0,2 (mol)

 2Al + 6HCl -> 2AlCl₃ + 3H₂

     0,2     0,6       0,2      0,3

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

=>mHcl=0,6.36,5=21,9g

=>mdd=219g