\(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
a, \(CaO+2HCl\rightarrow CaCl_2+H_2O\)
____0,1_____________0,1 (mol)
b, mCaCl2 = 0,1.111 = 11,1 (g)
c, m dd sau pư = 5,6 + 200 = 205,6 (g)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{11,1}{205,6}.100\%\approx5,4\%\)