Ca (ClO)2 + 4HCl → 2Cl2 + CaCl2 + 2H2O
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Ca (ClO)2 + 4HCl → 2Cl2 + CaCl2 + 2H2O
Mm Ca(ClO)2 = 143 g/molCl2 = 71 g/mol
V = 275 mL / 1000 = 0.275 L
g Cl2 a partir de Ca(ClO)2
g Cl2 = 50 g Ca(ClO2)2 x 1 mol x 2 mol Cl2 x 71 g Cl2
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143 g 1 mol Ca(ClO)2 1 mol Cl2
g Cl2 = 49,65
g HCl = 0.275 L x 6 mol HCl x 71 g Cl2
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11 L solc. 4 mol HCl 1 mol Cl2
g HCl = 58,58 g Cl2
Se forman 49,65 g Cl2 y sobra HCl.
cantidad sobrante: 0,275 x 6 moles iniciales – (49,65/71) x 2 = 0,25 moles = 0,25⋅36,5 = 9,18 g de HCl.