a)

$n_{Al} = 0,3(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$

b)

$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

c)

$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$

a) Mg + H₂SO₄ --> MgSO₄ + H₂

b) \(n_{Mg}=\dfrac{3}{24}=0,125\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

         0,125->0,125-->0,125-->0,125

=> V_H2 = 0,125.22,4 = 2,8 (l)

\(V_{dd.H_2SO_4}=\dfrac{0,125}{2}=0,0625\left(l\right)\)

c) Sản phẩm là Magie sunfat và khí hidro

\(m_{MgSO_4}=0,125.120=15\left(g\right)\)

m_H2 = 0,125.2 = 0,25 (g)

d) 

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,125}{1}\) => Hiệu suất tính theo H₂

Gọi số mol CuO bị khử là a (mol)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

               a--->a-------->a

=> 16 - 80a + 64a = 14,4 

=> a = 0,1 (mol)

=> n_H2(pư) = 0,1 (mol)

=> \(H=\dfrac{0,1}{0,125}.100\%=80\%\)

a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

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

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

=> V_CO2 = 0,2.22,4 = 4,48 (l)

b) \(m_{dd.CH_3COOH}=\dfrac{0,4.60}{12\%}=200\left(g\right)\)

c) m_{dd sau pư} = 21,2 + 200 - 0,2.44 = 212,4 (g)

=> \(C\%_{muối}=\dfrac{0,4.82}{212,4}.100\%=15,44\%\)

a)\(Na2CO3+H2SO4-->Na2SO4+CO2+H2O\)

\(n_{Na2CO3}=\frac{21,2}{106}=0,2\left(mol\right)\)

\(n_{CO2}=n_{Na2CO3}=0,2\left(mol\right)\)

\(V_{CO2}=0,2.22,4=4,48\left(l\right)\)

b)\(n_{Na2SO4}=n_{Na2CO3}=0,2\left(mol\right)\)

\(m_{Na2SO4}=0,2.142=28,4\left(g\right)\)

c)\(n_{H2SO4}=n_{Na2CO3}=0,2\left(mol\right)\)

\(V_{H2SO4}=\frac{0,2}{2}=0,1\left(l\right)\)

Na2CO3+H2SO4-->Na2SO4+H2O+CO2

0,2-----------0,2-----------0,2

nNa2CO3=21,2\106=0,2 mol

=>mNa2SO4=0,2.142=28,4 g

=>VH2SO4=10 l

n_Na2CO3=0,2(mol)

Na₂CO₃ + H₂SO₄ --> Na₂SO₄ + CO₂ + H₂O

n_CO2=n_Na2CO3=0,2(mol)<=>4,48 lít

n_Na2SO4=n_Na2CO3=0,2(mol)<=>142.0,2=28,4(g)

n_H2SO4=n_Na2CO3=0,2(mol)

=>V dd H2SO4=0,2/2=0,1(lít)

a, \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{FeSO_4}=n_{H_2}=n_{Fe}=0,01\left(mol\right)\)

\(\Rightarrow m_{FeSO_4}=0,01.152=1,52\left(g\right)\)

\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)

b, \(n_{H_2SO_4}=n_{Fe}=0,01\left(mol\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{0,01.98}{19,6\%}=5\left(g\right)\)

c, Ta có: m _{dd sau pư} = 0,56 + 5 - 0,01.2 = 5,54 (g)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{1,52}{5,54}.100\%\approx27,44\%\)

giup mk voi

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(0.2..........0.3...............0.1...........0.3\)

\(m_{H_2SO_4}=0.3\cdot98=29.4\left(g\right)\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{29.4\cdot100}{20}=147\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0.1\cdot342=34.2\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng }}=5.4+147-0.3\cdot2=151.8\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34.2}{151.8}\cdot100\%=22.53\%\)

\(n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\)

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

\(\dfrac{3}{14}....\dfrac{3}{14}.......\dfrac{3}{14}......\dfrac{3}{14}\)

\(m_{FeSO_4}=\dfrac{3}{14}\cdot152=32.57\left(g\right)\)

\(V_{H_2}=\dfrac{3}{14}\cdot22.4=4.8\left(l\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{\dfrac{3}{14}\cdot98}{19.6\%}=107.1\left(g\right)\)