\(n_{CaO}=\dfrac{11.2\cdot1000}{56}=200\left(kmol\right)\)
\(CaCO_3\underrightarrow{^{t^0}}CaO+CO_2\)
\(200...........200\)
\(m_{CaCO_3}=200\cdot100=20000\left(kg\right)=20\left(tấn\right)\)
$CaCO_3 \xrightarrow{t^o} CaO + CO_2$
n CaCO3 = n CaO = 11,2.1000/56 = 200(kmol)
m CaCO3 = 200.100 = 20 000(kg)
\(PTHH:CaCO_3\rightarrow CaO+CO_2\)
\(n_{CaO}=\dfrac{m}{M}=\dfrac{11200000}{56}=200000\left(mol\right)\)
Theo PTHH : \(n_{CaCO3}=n_{CaO}=200000\left(mol\right)\)
\(\Rightarrow m_{CaCO3}=n.M=20000000\left(g\right)=20\) tấn .
