Đáp án+Giải thích các bước giải:

 

Đáp án + Giải thích các bước giải:

$PTHH:$

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2(1)$

$2KClO_3\xrightarrow{t^o}2KCl+3O_2(2)$

$2KMnO_4+16HCl\to 2KCl+2MnCl_2+5Cl_2+8H_2O(3)$

$K_2MnO_4+8HCl\to 2KCl+MnCl_2+2Cl_2+4H_2O(4)$

$MnO_2+4HCl\to MnCl_2+Cl_2+2H_2O(5)$

$3Cl_2+6NaOH\xrightarrow{t^o}5NaCl+NaClO_3+3H_2O$

Ta có $m_{O_2}=m_{hh}-m_{CR}=39,5+24,5-51,36=12,64(g)$

$\Rightarrow n_{O_2}=\dfrac{12,64}{32}=0,395(mol)$

$n_{KMnO_4}=0,25(mol);n_{KClO_3}=0,2(mol);n_{NaOH}=5.0,3=1,5(mol)$

Theo PT(2): $n_{O_2(2)}=1,5n_{KClO_3}=0,3(mol)$

$\Rightarrow n_{O_2(1)}=0,395-0,3=0,095(mol)$

Theo PT(1): $n_{KMnO_4}=2n_{O_2}=0,19(mol);n_{MnO_2}=n_{K_2MnO_4}=0,095(mol)$

$\Rightarrow n_{K_2MnO_4(dư)}=0,25-0,19=0,06(mol)$

Theo PT(3+4+5): $\Sigma n_{Cl_2}=2,5n_{KMnO_4(dư)}+2n_{K_2MnO_4}+n_{MnO_2}$

$\Rightarrow \Sigma n_{Cl_2}=2,5.0,06+2.0,095+0,095=0,76(mol)$

Xét tỉ lệ: $n_{Cl_2}>\dfrac{n_{NaOH}}{2}\Rightarrow Cl_2$ dư

Theo PT: $n_{NaCl}=\dfrac{5}{3}n_{Cl_2}=\dfrac{19}{15}(mol);n_{NaClO_3}=0,5.n_{NaOH}=0,75(mol)$

$\Rightarrow m_Z=\dfrac{19}{15}.58,5+0,75.106,5=153,975(g)$