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CaCO3 = 100,08 g/mol
CO2 = 44,00 g/mol
CaCO3(s) + 2 HCl(ac) = CaCl2(aq) + CO2(g) + H2O(l)
100 g CaCO3 ------------ 44,00 g CO2
33,7 g CaCO3 ----------- ??
masa ( CO2) = 33,7 x 44,00 / 100 =
masa ( CO2) = 1482,8 / 100 =
massa ( CO2) = 14,828 gramos
44,00 g ------------- 22,4 L en TEP
14,828 g ----------- ??
V = 14,828 x 22,4 / 44,00
V = 332,1472 / 44,00
V = 7,5488 L