\(n_{CaCO_3}=\dfrac{500}{100}=5\left(g\right)\\ pthh:CaCO_3\underrightarrow{t^o}CaO+CO_2\uparrow\)
5 5
\(m_{CaO\left(lt\right)}=5.56=280\left(g\right)\\
H\%=\dfrac{182}{280}.100\%=65\%\)
CaCO3 -> CaO + CO2
nCaO = nCaCO3 = 500/100\(\dfrac{ }{ }\)= 5 (mol)
mCaOLt= 5x56 = 280 (g)
H = 182/280x100 = 65 %