10% là tạp chất
=> \(m_{CaCO_3}=\dfrac{1.90\%}{100\%}=0,9\left(kg\right)\)
\(\Rightarrow n_{CaCO_3}=\dfrac{0,9}{100}=\dfrac{9}{1000}\left(mol\right)\)
PTHH: \(CaCO_3-t^o->CaO+CO_2\)
Theo PT ta có: \(n_{CaCO_3}=n_{CaO}=\dfrac{9}{1000}\left(mol\right)\)
=> \(m_{CaO}=\dfrac{9}{1000}.56=0,504\left(kg\right)\)
=> \(H=\dfrac{0,45}{0,504}.100\%=89,28\%\)