En la reacción ajustada: 6 HCl 2 Fe → 2 FeCl3 3H2 a) ¿qué cantidad de HCl reaccionará con 10 g de Fe? b) ¿qué masa de FeCl3 y H2 se formarán? Datos Masas atómicas Fe = 55,85; H = 1; Cl=35,5 .
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6HCl + 2Fe → 2FeCl3 + 3H2
Mm HCl = 36.5 g/mol Fe = 55.85 g/mol FeCl3 = 162.35 g/mol H2 = 2 G/MOL
a) g HCl = 10 g Fe x 1 mol Fe x 6 mol HCl x 36.5 g HCl ```````````` ```````````````` ````````````````` 55.85 g Fe 2 mol Fe 1 mol HCl
g HCl = 19.60
b)
g FeCl3 = 10 g Fe x 1 mol Fe x 2 mol FeCl3 x 162.35 g FeCl3 ````````````` `````````````````` `````````````````````` 55.85 g Fe 2 mol Fe 1 mol FeCl3
g FeCl3 = 29.06
g H2 = 10 g Fe x 1 mol Fe x 3 mol H2 x 2 g H2
````````````` `````````````` ````````````` 55.85 g Fe 2 mol Fe 1 mol H2
g H2 = 0.537
Mm HCl = 36.5 g/mol Fe = 55.85 g/mol FeCl3 = 162.35 g/mol H2 = 2 G/MOL
a) g HCl = 10 g Fe x 1 mol Fe x 6 mol HCl x 36.5 g HCl ```````````` ```````````````` ````````````````` 55.85 g Fe 2 mol Fe 1 mol HCl
g HCl = 19.60
b)
g FeCl3 = 10 g Fe x 1 mol Fe x 2 mol FeCl3 x 162.35 g FeCl3 ````````````` `````````````````` `````````````````````` 55.85 g Fe 2 mol Fe 1 mol FeCl3
g FeCl3 = 29.06
g H2 = 10 g Fe x 1 mol Fe x 3 mol H2 x 2 g H2
````````````` `````````````` ````````````` 55.85 g Fe 2 mol Fe 1 mol H2
g H2 = 0.537
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