Question

Giúp mìn zới 😘🙆‍♀️

Answer 1

Ta có: \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 

CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O (1)

CaO + 2HCl ---> CaCl₂ + H₂O

a. Theo PT₍₁₎: \(n_{CaCO_3}=n_{CO_2}=0,15\left(mol\right)\)

=> \(m_{CaCO_3}=0,15.100=15\left(g\right)\)

=> \(\%_{m_{CaCO_3}}=\dfrac{15}{17,8}.100\%=84,27\%\%\)

\(\%_{m_{CaO}}=100\%-84,27\%=15,73\%\)

b. Ta có: \(m_{CaO}=17,8-15=2,8\left(g\right)\)

=> \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

Theo PT₍₁₎: \(n_{HCl}=2.n_{CaCO_3}=2.0,15=0,3\left(mol\right)\)

Theo PT₍₂₎: \(n_{HCl}=2.n_{CaO}=2.0,05=0,1\left(mol\right)\)

=> \(n_{HCl_{PỨ}}=0,3+0,1=0,4\left(mol\right)\)

Đổi 200ml = 0,2 lít

=> \(C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\)