\(m_{CaCO_3\left(bđ\right)}=500.80\%=400\left(g\right)\)
=> \(n_{CaCO_3\left(bđ\right)}=\dfrac{400}{100}=4\left(mol\right)\)
\(CaCO_3-^{t^o}\rightarrow CaO+CO_2\)
\(m_{cr}=78\%.500=390\left(g\right)\)
=> \(n_{CO_2}=\dfrac{500-390}{44}=2,5\left(mol\right)\)
Ta có: \(n_{CO_2}=n_{CaCO_3\left(pứ\right)}=2,5\left(mol\right)\)
=>\(H=\dfrac{2,5}{4}.100=62,5\%\)
