\(n_{HCl}=0,5\times1=0,5\left(mol\right)\)
CaO + 2HCl → CaCl2 + H2O
Theo PT: \(n_{CaO}=\frac{1}{2}n_{HCl}\)
Theo bài: \(n_{CaO}=\frac{2}{5}n_{HCl}\)
Vì \(\frac{2}{5}< \frac{1}{2}\) ⇒ HCl dư
Theo PT: \(n_{CaCl_2}=n_{CaO}=0,2\left(mol\right)\)
\(\Rightarrow m_{CaCl_2}=0,2\times111=22,2\left(g\right)\)
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nHCl = 0.5*1=0.5 mol
CaO + 2HCl --> CaCl2 + H2O
Bđ: 0.2_____0.5
Pư : 0.2_____0.4_____0.2
Kt: 0_______0.1______0.2
mCaCl2= 0.2*111=22.2g
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