\(CaCO_3\left(\frac{56250}{7}\right)\rightarrow CaO\left(\frac{56250}{7}\right)+CO_2\)
\(m_{CaCO_3}=1000000.90\%=900000\left(g\right)\)
\(n_{CaO}=\frac{450000}{56}=\frac{56250}{7}\)
\(\Rightarrow m_{CaCO_3\left(pứ\right)}=\frac{56250}{7}.100=\frac{5625000}{7}\left(g\right)\)
\(\Rightarrow H=\frac{\frac{5625000}{7}}{900000}.100\%=89,29\%\)