3) \(n_{NaOH}=0,12.1,5=0,18\left(mol\right)\)

PTHH: NaOH + HCl ---> NaCl + H₂O

            0,18----->0,18--->0,18

=> \(V_{ddHCl}=\dfrac{0,18}{2}=0,09\left(l\right)\)

=> \(V_{dd.sau.pư}=0,09+0,12=0,21\left(l\right)\)

=> \(C_{M\left(NaCl\right)}=\dfrac{0,18}{0,21}=\dfrac{6}{7}M\)

PTHH: 2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O

            0,18------>0,09

=> \(m_{ddH_2SO_4}=\dfrac{0,09.98}{20\%}=44,1\left(g\right)\)

=> \(V_{ddH_2SO_4}=\dfrac{44,1}{1,25}=35,28\left(ml\right)\)

Giả sử có 1 mol chất hữu cơ

=> \(n_C=6\left(mol\right)\)

BTNT C: \(n_{CO_2}=n_C=6\left(mol\right)\)

Ta có: \(n_{CO_2}:n_{H_2O}=1:1\Rightarrow n_{H_2O}=n_{CO_2}=6\left(mol\right)\)

BTNT H: \(n_H=2n_{H_2O}=12\left(mol\right)\)

Lại có: \(n_{O_2}=n_{CO_2}=6\left(mol\right)\)

BTNT O: \(n_{O\left(hchc\right)}+2n_{O_2}=2n_{CO_2}+n_{H_2O}\)

=> \(n_{O\left(hchc\right)}=6.2+6-6.2=6\left(mol\right)\)

Trong 1 mol hchc có \(\left\{{}\begin{matrix}n_C=6\left(mol\right)\\n_H=12\left(mol\right)\\n_O=6\left(mol\right)\end{matrix}\right.\)

Vậy CTHH của hợp chất là C₆H₁₂O₆

Đặt \(\left\{{}\begin{matrix}n_{KCl}=a\left(mol\right)\\n_{KBr}=b\left(mol\right)\end{matrix}\right.\)

PTHH: 

\(KCl+AgNO_3\rightarrow AgCl\downarrow+KNO_3\)

a------->a----------->a

\(KBr+AgNO_3\rightarrow AgBr\downarrow+KNO_3\)

b------->b----------->b

Do \(m_{\downarrow}=m_{AgNO_3}\)

=> 170a + 170b = 143,5a + 188b 

=> 26,5a = 18b

=> \(b=\dfrac{53}{36}a\)

=> \(\%m_{KCl}=\dfrac{74,5a}{74,5a+119b}.100\%=\dfrac{74,5a}{74,5a+119.\dfrac{53}{36}a}.100\%=29,84\%\)

=> \(\%m_{KBr}=100\%-29,84\%=70,16\%\)

Ta có: \(\left\{{}\begin{matrix}n_{KMnO_4\left(bđ\right)}=\dfrac{15,8}{158}=0,1\left(mol\right)\\n_{O_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\end{matrix}\right.\)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂ 

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

=> \(H=\dfrac{0,04}{0,05}.100\%=80\%\)