\(n_{CaO}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(CaCO_3\underrightarrow{t^0}CaO+CO_2\)
\(0.1..............0.1........0.1\)
\(m_{CaCO_3}=0.1\cdot100=10\left(g\right)\)
\(m_{CaCO_3\left(tt\right)}=\dfrac{10\cdot100}{90}=11.11\left(g\right)\)
\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)
CaCO3--- CaO+CO2 (1)
a,nCaO=5,6/56=0,1 mol
theo pt có
nCaCO3/nCaO=1/1
nCaCO3=0,1 mol
ADCT:m=n*M=0,1*100=10 g
b,theopt có
nCO2/nCaO=1/1
nCO2=0,1 mol
ADCT VCO2=n*22,4 =0,1 *22,4 =2,24 l