Ile g Ca(OH)2 mona otrzymac z 1 kg kalcytu o zawartosci 92 % CaCO3 ?(MCa = 40, MC = 12, MO = 16 g/mol)
CaCO3 + H2O -->Ca(OH)2 + CO21 mol 1 mol100g 74g920g y=680,8 g
92% - x100% - 1000gx=920g
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CaCO3 + H2O -->Ca(OH)2 + CO2
1 mol 1 mol
100g 74g
920g y=680,8 g
92% - x
100% - 1000g
x=920g