\(PTHH:KClO_3+HCl\rightarrow3H_2O+KCl+3Cl_2\left(1\right)\)
\(2Fe+3Cl_2\rightarrow2FeCl_3\left(2\right)\)
Ta có :
\(n_{Fe}=\frac{16,25}{56}=0,3\left(mol\right)\)
Theo PT2: \(\Rightarrow n_{Cl2}=0,45\left(mol\right)\)
Theo PT1: \(\Rightarrow n_{KClO3}=0,15\left(mol\right)\)
\(\Rightarrow m_{KClO3}=0,15.122,5=18,375\left(g\right)\)
Theo PT1: \(n_{HCl}=0,15\left(mol\right)\)
\(\Rightarrow V_{HCl}=\frac{0,15}{8}=0,01875\left(l\right)=18,75\left(ml\right)\)
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