a) 2NaOH + Cl₂ --> NaCl + NaClO + H₂O

b) \(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: 2NaOH + Cl₂ --> NaCl + NaClO + H₂O

              0,1<----0,05--->0,05----->0,05

=> \(V_{ddNaOH}=\dfrac{0,1}{1}=0,1\left(l\right)\)

c) 

\(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,05}{0,1}=0,5M\\C_{M\left(NaClO\right)}=\dfrac{0,05}{0,1}=0,5M\end{matrix}\right.\)

\(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)

 2 mol        1 mol    1mol      1mol      1 mol

0,1          0,05         0,05         0,05      0,05

\(V_{NaOH}=\dfrac{0,1}{1}=0,1\left(l\right)\)

\(CM_{NaCl}=\dfrac{0,05}{0,1}=0,5M\)

\(CM_{NaClO}=\dfrac{0,05}{0,1}=0,5M\)

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

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:    0,1         0,1            0,1          0,1

\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

\(C_{M_{ddH_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\)

\(C_{M_{ddFeSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

\(n_{Cl_2}=\dfrac{0.0448}{22.4}=0.002\left(mol\right)\)

\(n_{NaOH}=0.05\cdot1=0.05\left(mol\right)\)

\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)

\(2.................1\)

\(0.05........0.002\)

LTL : \(\dfrac{0.05}{2}>\dfrac{0.002}{1}\)

NaOH dư

\(C_{M_{NaCl}}=C_{M_{NaClO}}=\dfrac{0.002}{0.05}=0.04\left(M\right)\)

\(C_{M_{NaOH\left(dư\right)}}=\dfrac{0.05-0.002\cdot2}{0.05}=0.92\left(M\right)\)

a) $2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
b) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
Theo PTHH : $n_{Cl_2} = \dfrac{3}{2}n_{Al} =  0,3(mol)$
$\Rightarrow V_{Cl_2} = 0,3.24,79 = 7,437(lít)$

c) $n_{AlCl_3} = n_{Al} = 0,2(mol)$
$\Rightarrow m_{AlCl_3} = 0,2.133,5 = 26,7(gam)$

\(n_{Na_2S_2O_3}=0.025\cdot0.1=2.5\cdot10^{-3}\left(mol\right)\)

\(2Na_2S_2O_3+I_2\rightarrow2NaI+Na_2S_4O_6\)

\(2.5\cdot10^{-3}....1.25\cdot10^{-3}\)

\(2KI+Cl_2\rightarrow2KCl+I_2\)

\(2.5\cdot10^{-3}............1.25\cdot10^{-3}\)

\(C_{M_{KI}}=\dfrac{2.5\cdot10^{-3}}{0.2}=0.0125\left(M\right)\)

19,5g gì :) ?, với cả đề cho V_dd thì sao tính đuợc C% :))