PTHH: CaO + 2HCl -> CaCl2 + H2O
Ta có: \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ =>n_{HCl}=0,1.2=0,2\left(mol\right)\\ =>m_{HCl}=0,2.36,5=7,3\left(g\right)\\ =>m_{ddHCl}=\dfrac{100.7,3}{14,6}=50\left(g\right)\)
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