Ta có: \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
a. PTHH; CaO + 2HCl ---. CaCl2 + H2O
b. Theo PT: \(n_{HCl}=2.n_{CaO}=2.0,05=0,1\left(mol\right)\)
Đổi 200ml = 0,2 lít
=> \(C_{M_{HCl}}=\dfrac{n_{ct_{HCl}}}{V_{dd_{HCl}}}=\dfrac{0,1}{0,2}=0,5\)(mol/l)
c. Theo PT: \(n_{CaCl_2}=n_{CaO}=0,05\left(mol\right)\)
=> \(m_{CaCl_2}=0,05.111=5,55\left(g\right)\)