Question

Answer 1

$m_{CaCO_3} = 1,5.1000.80\% = 1200(gam)$

$CaCO_3 \xrightarrow{t^o} CaO + CO_2$
$n_{CaCO_3\ pư} = n_{CaO} = \dfrac{537,6}{56} = 9,6(kmol)$

$H = \dfrac{9,6.100}{1200}.100\% = 80\%$

 

Answer 2

Ta có:1,5 tấn = 1500 kg

⇒ \(m_{CaCO_3}=1500.80\%=1200\left(kg\right)\)

\(n_{CaO}=\dfrac{537,6}{56}=9,6\left(mol\right)\)

\(PTHH:CaCO_3\underrightarrow{t^o}CaO+CO_2\)

Mol:           9,6            9,6

⇒\(H=\dfrac{9,6.100}{1200}.100\%=80\%\)