Question

In urma reactiei Hcl cu CaCo3 se degaja 44,8 litri Co2
a)mCaCo3=?
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Answer 1

    
                       x g                       88 g
           2HCl +CaCO3 =CaCl2 +CO2 +H2O
                     100 g                      44 g


Se transforma volumul de CO2 in moli de CO2

44,8:22,4=2 moli CO2
n=m:M                                                  n=nr.de moli
m=n. M                                                M= masa moleculara
m=2.44=88 g CO2                               m =cantit . de substanta


x=100.88:44=200 g CaCO3


MCaCO3=40+12+3.16=100
MCO2=12+2.16=44