Ta có nCaO = \(\dfrac{56}{56}\) = 1 ( mol )
nCO2 = \(\dfrac{44}{44}\) = 1 ( mol )
nCaCO3 = \(\dfrac{100}{100}\) = 1 ( mol )
CaO + CO2 \(\rightarrow\) CaCO3
1.............1.............1
=> nCO2 = \(\dfrac{22}{44}\) = 0,5 ( mol )
CaO + CO2 \(\rightarrow\) CaCO3
0,5..........0,5.........0,5
=> mCaO = 56 . 0,5 = 28 ( gam )
=> mCaCO3 = 100 .0,5 = 50 ( gam )
Đúng 0
Bình luận (0)